OUP -24 -4-4-69—5,01 OSMANIA UNIVERSITY LIBRARY Gall No. 5 / X/ & S& n Accession No. Q % ' S °~ ' Author £^''<vt^^y?Ajro^fc<£ ^y/u^ef/ZToh^/i This b<#>k should be returned on or before the date last marked below. Algebra and Trigonometry by ALVIN K. BETTINGER Head of Department and Professor of Mathematics The Crei pli ton University and JOHN A. ENGLUND Formerly Assistant Professor of Mat lie mattes The Cr eight on University INTERNATIONAL TEXTBOOK COMPANY Scranton, Pennsylvania International Textbooks in Mathematics L R. Wilcox Professor of Mathematics Illinois Institute of Technology Consulting Editor Second Printing, ] anuary, 1963 Copyright © I960, by International Textbook Company. All rights reserved. Printed in the United States of America by The H addon Craftsmen } Inc., at Scr anion, Pennsylvania. Library of Congress Card Number: 60-9987. Preface The authors believe that in this book the basic material of college algebra and trigonometry has been presented with suflicient rigor to provide a firm and coherent groundwork for subsequent courses in mathematics. The material is presented in such a way that it can be grasped by the student without undue assistance. The first chapter consists of a number of introductory topics which are intended to serve as a review of elementary algebra. Actually, something more than a mere review is available in this chapter. Not only are the review topics considered from a more mature point of view than is usual, but the treatment is inter- woven with concepts that are basic for an understanding of more advanced mathematical topics. We begin with the algebra of the real-number system. Axioms pertaining to fundamental operations are given, and the various rules for the elementary operations of algebra are derived and logically connected with the basic assump- tions. We are led naturally to an ordering of the real-number system and to the foundation for a later chapter on inequalities that is easier to understand and more useful than the treatment one customarily finds in textbooks. The second chapter introduces the student to the function con- cept, which serves as a basis for much of the remaining work of the book. Certain aspects of the discussion become somewhat abstract, but the student is reminded that a proper understanding of the true nature of a function is important for virtually all later courses in mathematics. In line with modern demands, the trigonometric functions are initially introduced in the third chapter as functions of real num- bers. Following this presentation, the transition to functions of angles is relatively simple. The rest of the volume contains all the usual topics from college algebra and trigonometry. In certain instances a particular devel- opment may differ somewhat from that usually found. In such cases, the authors believe, the departure is to the advantage of the student. vi Preface We are indebted to our colleague, Professor Morris Dansky, for his valuable suggestions while the manuscript was in preparation. We wish particularly to express our deep appreciation to Pro- fessor L. R. Wilcox for his thorough criticism of the manuscript and his invaluable suggestions for improvement of the text. Finally, a special word of thanks is due the International Textbook Com- pany for its cooperation and patience. A. K. Bettinger J. A. Englund Omaha, Nebraska August, 1960 Contents 1. Introductory Topics 1 1-1. The Real-Number System 1 1-2. Fundamental Assumptions 1 1-3. Operations With Zero 5 1-4. Reciprocals 5 1-5. The Real-Number Scale 6 1-6. Rules of Signs 7 1-7. Fundamental Operations on Fractions 9 1-8. Order Relations for Real Numbers 12 1-9. Absolute Value 14 1-10. Inequalities Involving Absolute Values 14 1-11. Positive Integral Exponents 16 1-12. Algebraic Expressions 17 1-13. Equations and Identities 18 1-14. Symbols of Grouping 18 1-15. Order of Fundamental Operations 20 1-16. Addition and Subtraction of Algebraic Expressions 21 1-17. Multiplication of Algebraic Expressions 22 1-18. Special Products 22 1-19. Division of Algebraic Expressions 23 1-20. Factoring 25 1-21. Important Type Forms for Factoring 27 1-22. Greatest Common Divisor 30 1-23. Least Common Multiple 32 1-24. Reduction of Fractions 33 1-25. Signs Associated With Fractions 34 1-26. Addition and Subtraction of Fractions 36 1-27. Multiplication and Division of Fractions 39 1-28. Complex Fractions 40 1-29. Linear Equations 42 1-30. Linear Equations in One Unknown 43 2. The Function Concept 49 2-1. Rectangular Coordinate Systems in a Plane 49 2-2. Distance Between Two Points 50 2-3. Functions 52 2-4. Functional Notation 55 2-5. Some Special Functions 57 2-6. Variation 57 2-7. Classification of Functions 61 viii Contents 3. The Trigonometric Functions G3 3-1. The Point Function P(t) 63 3-2. Definitions of the Trigonometric Functions (54 3-3. Identities 68 3-4. Tables of Trigonometric Functions 71 3-5. Positive and Negative Angles and Standard Position 75 3-6. Measurement of Angles 76 3-7. The Relation Between Radians and Degrees 77 3-8. Arc Length and Area of a Sector 78 3-9. Trigonometric Functions of Angles 81 3-10. Tables of Natural Trigonometric Functions of Angles 82 4. The Laws of Exponents 86 4-1. Positive Integral Exponents 86 4-2. Meaning of a 88 4-3. Negative Exponents 90 4-4. Scientific Notation 1)2 4-5. Rational Exponents 1)2 4-6. The Factorial Symbol 1)7 4-7. The Binomial Theorem 1)7 4-8. General Term in the Binomial Expansion 99 5. Logarithms 101 5-1. Definition of a Logarithm 101 5-2. Laws of Logarithms 102 5-3. Systems of Logarithms 105 5-4. Common Logarithms 105 5-5. Rules for Characteristic and Mantissa 106 5-6. How to Write Logarithms 108 5-7. How to Use a Table of Mantissas 108 5-8. Logarithmic Computation 110 5-9. Change of Base 113 6. Right Triangles and Vectors 115 6-1. Rounding Off Numbers 115 6-2. Trigonometric Functions of Acute Angles 116 6-3. Procedures for Solving Right Triangles 117 6-4. Angles of Elevation and Depression 120 6-5. Bearing in Navigation and Surveying 121 6-6. Projections 122 6-7. Scalar and Vector Quantities 125 6-8. Logarithms of Trigonometric Functions 131 6-9. Logarithmic Solution of Right Triangles 133 7. Trigonometric Functions of Sums and Differences 135 7-1. Derivation of the Addition Formulas 135 7-2. The Double-Angle Formulas 139 7-3. The Half-Angle Formulas 140 7-4. Products of Two Functions Expressed as Sums, and Sums Expressed as Products 143 Contents ix 8. Graphs of Trigonometric Functions; Inverse Functions and Their Graphs 146 8-1. Variation of the Trigonometric Functions 140 8-2. The Graph of the Sine Function 117 8-3. The Graphs of the Cosine and Tangent Functions 148 8-4. Periodicity, Amplitude, and Phase 14i) 8-5. Inverse Functions 155 8-6. Inverses of the Trigonometric Functions 156 9. Linear Equations and Graphs 163 9-1. Solutions of Simultaneous Equations 163 9-2. Algebraic Solution of Linear Equations in Two Unknowns... 166 9-3. Linear Equations in Three Unknown* 1(57 9-4. Graphs of Linear Functions 169 9-5. Intercepts 170 9-6. Graphical Solution of Linear Equations in Two Unknowns . . . 171 10. I JETERMINANTS 173 10-1 . I )eterminants of the Second Order 173 10-2. Determinants of the Third Order 175 10-3. Properties of Determinants 177 10-4. Solution of Three Simultaneous Linear Equations in Three Unknowns 181 10-5. Systems of Three Linear Equations in Three Unknowns When D = () 183 10-6. Homogeneous Equations 184 10-7. Sum and Product of Determinants 185 1 1 . Complex Numbers 189 The Complex Number System 189 11-1 11-2, 11- 11-4. 11-5 11-6 11-7. 11-8, The Standard Notation for Complex Numbers 191 Operations on Complex Numbers in Standard Form 192 Graphical Representation 195 Trigonometric Representation 196 Multiplication and Division in Trigonometric Form 198 Dc Moivre's Theorem 199 Roots of Complex Numbers 200 12. Equations in Quadratic Form 204 12-1. Quadratic Equations in One Unknown 204 12-2. Solution of Quadratic Equations by Factoring 20 4 12-3. Completing the Square 206 12-4. Solution of Quadratic Equations by the Quadratic Formula . . . 209 12-5. Equations Involving Radicals 212 12-6. Equations in Quadratic Form 21 1 12-7. The Discriminant 215 12-8. Sum and Product of the Roots 217 12-9. Graphs of Quadratic Functions 218 12-J 0. Quadratic Equations in Two Unknowns 221 x Contents 12-11. Graphical Solutions of Systems of Equations Involving Quadratics 224 12-12. Algebraic Solutions of Systems Involving Quadratics 227 12-13. Exponential and Logarithmic Equations 231 12-14. Graphs of Logarithmic and Exponential Functions 233 13. Theory of Equations 235 1 3-1. Introductory Remarks 235 13-2, Synthetic Division 235 13-3. The Remainder Theorem 240 13-4. The Fundamental Theorem of Algebra 241 13-5. Pairs of Complex Roots of an Equation 243 13-6. The Graph of a Polynomial for Large Values of a- 244 13-7. Roots Between a and b If f(a) and f(b) Have Opposite Signs 245 13-8. Rational Roots 245 14. Inequalities 248 14-1. Introduction 248 14-2. Properties of Inequalities 248 14-3. Solution of Conditional Inequalities 249 14-4. Absolute Inequalities 254 15. Progressions 256 15-1 . Sequences and Series 256 15-2. Arithmetic Progressions 260 15-3. The General Term of an Arithmetic Progression 260 15-4. Sum of the First n Terms of an Arithmetic Progression 261 15-5. Arithmetic Means 262 15-6. Harmonic Progressions 264 1 5-7. Geometric Progression 265 15-8. The General Term of a Geometric Progression 265 15-9. Sum of the First n Terms of a Geometric Progression 266 15-10. Geometric Means 267 15-11. Infinite Geometric Progression 268 15-12. Repeating Decimals 269 15-13. The Binomial Series 271 16. Mathematical Induction 273 16-1. Method of Mathematical Induction 273 16-2. Proof of the Binomial Theorem for Positive Integral Exponents 275 17. Permutations, Combinations, and Probability 278 17-1. Fundamental Principle 278 17-2. Permutations 279 17-3. Permutations of n Things Not All Different 280 17-4. Combinations 281 17-5. Binomial Coefficients 282 17-6. Mathematical Probability 283 17-7. Most Probable Number and Mathematical Expectation 284 17-8. Statistical, or Empirical, Probability 284 Contents xi 17-9. Mutually Exclusive Events 285 17-10. Dependent and Independent Events 286 17-11. Repeated Trials 287 18. Solution of the General Triangle 289 18-1. Classes of Problems 289 18-2. The Law of Sines 289 18-3. Solution of Case I by the Law of Sines: Given One Side and Two Angles 290 18-4. Solution of Case II by the Law of Sines: Given Two Sides and the Angle Opposite One of Them 291 18-5. The Law of Cosines 296 18-6. Solution of Case III and Case; IV by the Law of Cosines 297 18-7. The Law of Tangents 298 18-8. The Half-Angle Formulas 300 18-9. Area of a Triangle 302 Appendix A. Tables 307 B. Answers to Odd-Numbered Problems 337 Index 353 1 Introductory Topics 1-1. THE REAL-NUMBER SYSTEM The real-number system that we use in the early part of this course is a development from the original counting numbers, or positive integers, such as 1, 2, and 3. Almost simultaneously with the invention of positive integers, practical problems of measure- ment gave rise to positive fractions, such as 1/2, 5/6, and 16/7. Much later, in comparatively modern times, the concepts of negative numbers and of other types of numbers were gradually developed. Negative numbers were invented when the problem of subtracting one number from a smaller one presented itself. Thus, the number system was soon enlarged to include the negative integers and fractions. These positive and negative numbers, together with zero, are called the rational numbers. Hence, a rational number is defined to be any number that can be expressed as the quotient, or ratio, of two integers. For example, —2/3, 5 (which may be con- sidered as 5/1), and —7 are rational numbers. The number system was then extended to include also numbers which cannot be expressed as the quotient of two integers, namely, the irrational n}nnbers; examples are y/2 and tt. The two classes of numbers, rational and irrational, comprise the real members. These numbers are so called in contrast to the imaginary or complex numbers considered in Chapter 11. 1-2. FUNDAMENTAL ASSUMPTIONS We shall proceed to introduce the four fundamental operations of addition, subtraction, multiplication, and division into the system of real numbers. The reader has probably been performing these operations in arithmetic and algebra without being conscious that certain basic laws were being obeyed. We shall introduce the four fundamental operations and state, without proof, the laws or assumptions governing them. l 2 Introductory Topics Sec. 1-2 Addition. It is assumed that there is a mode of combining any two real numbers a and 6 so as to produce a definite real number called their sum. This mode of combination is called addition. The sum of a and b is denoted by a + b. In this sum a and b are called terms. Multiplication. It is assumed that there is a mode of combining any two real numbers a and b to produce a definite real number called their product. This mode of combination is called multiplica- tion. The product of a and b is denoted by a • b or by ab. The individ- ual numbers a and b are called factors of the product. Commutative Law for Addition. If a and b are any real numbers, then (1-1) a + b=b + a. Thus 1 , the sum of two numbers is the same regardless of the order in which they are added. For example, 2 + 3=3 + 2. Associative Law for Addition. If a, b, c are any real numbers, then (1-2) (a + b) + c = a + (b + c). That is, we obtain the same result whether we add the sum of a and 6 to c, or we add a to the sum of b and c. Since the way in which we associate or group these numbers is immaterial, we may write this common value as a + 6 4- c without fear of ambiguity. For example, 2 + 3 + 4 = (2 + 3) + 4 = 2 + (3 + 4). Commutative Law for Multiplication. If a and b are any real num- bers, then (1-3) ab = ba. That is, the product of two numbers is the same regardless of the order in which they are multiplied. For example, 2-3 = 3-2. Associative Law for Multiplication. , If a, b, c are any real num- bers, then (1-4) (ab)c = a(bc). 1 Illustrations of the laws are given here only for the most familiar num- bers, the positive integers. It is understood, however, that the laws apply to all real numbers. Sec. 1—2 Introductory Topics 3 That is, we obtain the same result whether we multiply the product of a and b by c, or we multiply a by the product of b and c. Since the way in which we associate or group these numbers is imma- terial, we may write the result as abc without fear of ambiguity. Thus 2 • 3 • 4 = (2 • :}) ■ l = 2 • (3 • 4). Distributive Law. If a, b, c are any real numbers, then 2 (1 , r >) a(b + c) = ub + ac. This law, which is usually known as the distributive law for multi- plication with respect to addition, effects a connection between addition and multiplication. The distributive law forms the basis for the factoring process in algebra, as will be seen. A simple example of the distributive law is 2- (3 + 4) - 2-3 + 2. 1. This law can be extended to the case where the sum consists of three or more terms, as in the following illustration : :\a(x + 2// - 3;) = liax + 0u// - Uaz. For positive integers, multiplication may also be interpreted as repeated addition. Thus, by the distributive law, 3 • 4 = (1 + 1 + 1) • 1 -- (1 • 4) + (1 • 4) + (1 • 4) = 4 + 4 + 4, 3-4 = 3(1 + 1 + 1 + 1) = (3- 1) + (3- 1) + (3- I) + (3- 1) = <*> + - + - + 3. Zero. It is assumed that there is a special number called zero and denoted by 0, such that, for every real number a, (1 (i) a + = a. For example, 3 + 0=3, 0+1=1, + = 0. It can be easily shown that only one number with the property of can exist. For let 0' be another such number. Then, since a + = a and b + 0' = b for any numbers a, b, it follows, by taking a -— 0' and b ~ 0, that ()' + = 0', and + 0' = 0. From the commutative law, — 0'. 2 The right side of (1-5) should read (ah) -f (ac). However, by conven- tion, we agree to omit the parentheses when all multiplications are to be performed before any addition. 4 Infroductory Topics Sec. 1—2 Negative of a Number. It is assumed that for every real number a there exists a corresponding number, called the negative of a and designated by —a, such that (1 7) a + (- „) = 0. For example, 1 + (- 1) =0, (- 2) + 2 =0. That each number has but one negative may be shown in the following way: Let x be another negative of a, so that a + x — 0. Then -a = (-<i)+0 = (-„) + (<* + s). By associativity, - a = ((- a) + a) + x, or — rt = + x — x. In particular, the negative of zero is - = - + = 0. The Unit. It is assumed that there is a special number called the unit and denoted by 1, such that, for every real number a, (1 8) a • 1 = a. There cannot be a second unit 1'. If there were, we could say that 1 ■ 1' = 1, 1'. 1 = r, whence 1 = 1'. Reciprocal of a Number. It is assumed that for every number a which is not 0, there is an associated number - , called the rccip- a vocal of a, such that (1-9) «•- = 1. a The reader may verify the fact that there is only one reciprocal of each number. Thus, if x is another reciprocal of a, that is, if a • x — 1, then x = - • a It is important to note the restriction a ^ in the definition of the reciprocal. In the next section we shall see why this restriction is needed. Subtraction. The difference a — b, of any real numbers a and b, is defined by (1 10) a - b = a + (- 6). Sec. 1—4 Introductory Topics 5 The operation indicated by the si#n minus which produces for any two real numbers a and b the real number a — b is called subtraction. Division. The quotient a f b or 7 or a -r- fr of any real numbers a b and fr, where b ^ 0, is defined bv a-.., ; = «.©• The operation associating with real numbers a and fr (fr -V 0) their quotient is called division. It should be noted that subtraction and division are subordinate to addition and multiplication, in that they are defined in terms of these latter. The difference a — b is that number .r for which b + .v - a. Also, the quotient u b is that number \j for which b • y = «. It should be noted that a — a — a + (-«) = for every number a, and that r/ 7/ - a • (1, a) — 1 for every number a y 0. 1-3. OPERATIONS WITH ZERO It has already been noted that the special number has the prop- erty a + - a for every real number a. In particular, we may let a = to obtain () + () = 0. It has already been noted that —0 = 0, so that a — a 4- = a — for every real number a. Next, we prove that for every real number a, (1-12) a •() -r (). Let # — ft • 0. Then, by the distributive law, .r = « • = a • (0 + 0) = f/ • + a • - x + x. If we add — .r, we obtain - :r + (- jc) = U + x ) + (- x ) = x + Or + (- x)) = x + = x. Since # = a • 0, (1-12) is established. From this last result, it follows that, for b i- 0, (113) l = °-(i)=<>- 1-4. RECIPROCALS It was noted that every non-zero number has a reciprocal. We can now see why cannot have a reciprocal. If has a reciprocal x, then • x = 1. Since it has been shown that • x = 0, we would have 6 Introductory Topics Sec. 1-4 to conclude that = 1. However, if = 1 is allowed, then for every number a we have a = 1 • a — • a — 0. Hence, would be the only number in the number system. This situation obviously should be ruled out. Therefore, cannot have a reciprocal. Moreover, since a/b = a • (1/6), the quotient a/b is not defined when 6 = 0. The reciprocal of the product of two non-zero numbers can be expressed in another way : 1 1 1 (1-14) a • b ab provided that neither a nor b is 0. To prove this result, we begin with 11 7 1 l K , 1 1 - • i •a«6=--a»T*o= l • 1 =1. a b a b We then multiply by — r to obtain 1 1 111 1111 a • b a * b a b a • b a b a b In the preceding proof, free use has been made of the commutative and associative laws. Finally, if a ^ 0, the reciprocal of the reciprocal of a is a itself. Thus, (1-15) ^ = a. Since I7ri -a/«) = i. multiplication by f/ gives a ~ 1 • a = -7- • (1/a) • a = — r- • 1 = 77 - • 1/a 1/a 1/a 1-5. THE REAL-NUMBER SCALE Real numbers may be represented by points on a straight line. On such a line select an arbitrary point as origin and lay off equal unit distances in both directions, as shown in Fig. 1-1. (The — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — 1 — -7 _ 6 -5 -4 -3 -2 -1 I 2 3 4 5 6 7 Fiu. 1-1 unit segment may have any length whatsoever.) Label the points thus far specified as indicated : is the origin, 1 is the first point to Sec. 1—6 Introductory Topics 7 the right, 2 is the second point to the right, —1 is the first point to the left, and so on. Rational numbers that are not integers cor- respond to certain other points in a natural way. For example, 1/2 corresponds to the midpoint of the segment joining points labeled and 1; and —7/3 represents the point one-third of the dis- tance from the point —2 to the point —3. It is a basic assumption concerning the real numbers that every point corresponds to a unique real number, and that every real number corresponds to exactly one point. The full significance of this assumption cannot be developed in an elementary text. One observation of importance can be made at this time. The non-zero real numbers are divided into two classes. One class con- sist of numbers representing points to the "right" of 0, and the other consists of numbers representing points to the "left" of 0. The first class consists of positive numbers, and the second of negative numbers. The number may be considered as constituting a third class. It is understood that no two of the three classes — zero, positive numbers, and negative numbers — have any numbers in common. Thus a number cannot be both positive and zero, both positive and negative, or both negative and zero. The specific desig- nation of any negative number will include an explicit sign — , which is prefixed. (This convention, however, does not exclude the possi- bility of allowing a general symbol, such as x, to stand for a nega- tive number.) Positive numbers do not require such a sign, although frequently the sign f is used. It is to be assumed that the sum of two positive numbers is positive, as is also the product of two positive numbers. 1-6. RULES OF SIGNS To operate effectively with real numbers, a knowledge of the rules of signs and of properties of negative numbers is essential. In each of the following relationships, a and b are any two real numbers, except that the denominator of a fraction may not be zero. (1-16) - (- a) = a. (1-17) - ( a + b) = - a - b. (1-18) - ( a -b) = - a + b. (1-19) (- a)b = — (ab); in particular, (— \)b = - b (1-20) (-«)(- b) =ab. (1-21) 1 1 - 6 ~ b' 8 Introductory Topics Sec. 1-6 Proofs of (1-16) to (1-28): (1-16) - (- a) = - (- a) = a + (- a) + (- (- a)) = a + = a. (1-17) - (a + 6) = + - (a + 6) = - a + a + (- b) + b - (a + b) = - a-b+ (a + b) - (a + b) = -a-b + = -a-b by (1-6), (1-7). (1-18) - (a - b) = - (a + (- 6)) = - a - (- b) = - a + 6 by (1-17), (1-16). (1-19) Since (-a)-b + a-6 = (-a + a)-6 = 0-6 = by (1-12), (- a) • 6 = (- a) • 6 + a • 6 - (a • 6) = - (a • 6) = - (a • 6). (1-20) . (- a) (- 6) = - ((- o) • b) = - (- (c • 6)) = ab by (1-19), (1-16). d-21) ^ = ^.l=^.6.1=J.. ( - 6) .(.l) = l-(-|) = -£ ^ (1-20). ('■*> :h = -^ = -(-i) = -(-i)--ii ^£ = (. a) .l = .(a.l) = -J by (1-19). (1-23) JZ^ = _ (_ |) = « by (1-22), (1-16). , It has already been observed that non-zero numbers are divided into two classes, namely, positive and negative. It is assumed that if & is positive, then -a is negative ; and that if a is negative, then —a is positive. All calculations involving negative numbers can be made by performing calculations with positive numbers and apply- ing one or more relationships just given. It follows from (1-17), for example, that the sum of two negative numbers is negative, and is equal to the negative of the sum of the negatives of the given numbers. Also, from (1-20) it follows that the product of two negative numbers is positive, and is equal to the product of the negatives of the given numbers. By (1-19) the product of a posi- tive number and a negative number is negative. Sec. 1—7 Introductory Topics 9 1-7. FUNDAMENTAL OPERATIONS ON FRACTIONS A further study of the algebra of real numbers leads us to the consideration of the fundamental operations as applied to fractions. By definition, a fraction is the quotient obtained by dividing one number a by another number 6, where b is not zero. We call a the numerator and b the denominator; and we generally write the fraction a/6, read "a over b" or "a divided by b" We shall list the following basic relationships for applying the four operations to fractions. In them a, 6, c, d are any real num- bers, except that no factor in the denominator of a fraction may be zero. ac __ a bc~b* (1-24) (1-25) a + b^a + b c c (1-26) a - + I = c d a __ b c c a __b c ~d~" cd c ad + be cd a — b c ad — be (1-27) (1-28) (1-29) (i-30) ?/£ = l- d - = f 7 b / d b e be A special case of (1-30) is a c _ ac b'd~W' 1 / c a /d = ~c' which states that the reciprocal of a fraction is found by inverting the fraction. Also, by (1-30), dividing by a fraction is equivalent to multiplying by its reciprocal. Proofs of (1-24) to (ISO): (1-29) Since y^ = Td b ^- 14 >' a c b'd '' 1 1 i = a t • c • 3 = (a • b d v \ 1 1 = a • 1 c 'bd~ ac bd • 0-24) ac _ a c be ~ b c = ?•! = a 6 by (1-29). (1-25) 5 + b 1 , , 1 c c c = (0 + 6) 1 c a + b c .by (1-5). 10 Introductory Topics Sec. 1-7 «- 26 > ° + 3 = ^ + ^ = ^ » y(1 -24),(l-25>. (1-27) = - + = a • - + (- 6) • - = (a - 6) - a — 6 by (1-22), (1-5). (1 _28) 2 » = ? + hfi = i^ + hi^ c d c a c • d c • d ad — be «■*> i/j-H-r;- a 1 a ad ad ad by (1-22), (1-24), (1-27). bl d be T c i c 1 , cd be -, b • -= 6 • i • a o d d d I'd a d by (1-29), (1-24). Example 1-1. State which of the fundamental assumptions are employed in each of the following equations: a) 3 +9 + (-5) =3 + (-5) +9. b) 11 + (6 + 3) + 7 = (11 + 6) + (3 + 7). c) (2 • 3) • 5 = 2 • (3 • 5). d) 5(3 + 4) = 5 • 3 + 5 • 4. Solution: a) The associative and commutative laws for addition. b) The associative law for addition. c) The associative law for multiplication. d) The distributive law. Example 1-2. In each of the following, perform the indicated operation: a) 2+3. b) (-2) +(-3). c) 5 + (-3). d) (+5) -(+3), e) (-7) -(+6). /) (-7) -(-6). Solution: Each case can be treated as an addition. a) 2 + 3 = 5. b) (-2) +(-3) = -5. c) 5 + (-3) =2. d) (+ 5) ~(+3) = +5 + (-3) = +2. e) (-7) -(+6) = -7 + (-6) = -13. /) (-7) -(-8) = -7 + <+6) = -1. Example 1-3. By using the fundamental assumptions and rules for operations* and transforming the left side into the right side, justify the equation (a + 6) - (c - d) = (6 - c) + (a + d). Sec, 1-7 Introductory Topics 1 1 Solution: By (1-18), (a + 6) - (c - d) = (a + 6) - c + d. By (1-1) and (1-2), (a + 6) - c + d = (6 - c) + (a + d). Example 1-4. By using the fundamental assumptions and rules of operations, justify the equation 6 ^ ac _ be a b — c ~ 6 — c Solution: By (1-29) and (1-24), b ac __ a • (6c) _ 6c a 6— c — a* (6— c) ~~ 6 — c EXERCISE 1-1 1. Identify the fundamental law or laws that justify each of the following equations: a. x + y — y + x. b. rs = sr. c. 2(3 • 5) = (2 • 3)5. d. 5(a + 6) = 5a + 56. e. (a -f 2) (6 - 3) = (6 - 3) (a + 2). f. (a + 6)c = c(a + 6) = ca + cb. 2. Find the value of each of the following: a. (-3) +(+5). b. (-5) +(-3). c. (-1) -(-2). d. (+7) -(+2). e. (-8) -(-9). f. 0-(-2). g. (- 5) - 0. h. 15 + (- 3). i. (- 7) - (- 5). j. (-5) -(+5). k. + (-3) -(+4). 1. (+32) -(-23) +(-45). 3. Evaluate each of the following: a. (+2) (-3). b. (-3) (-5). c. (-7) (+5). d. (-5) (-9). e. 2(-5). f. (-7)0. g. (-l)(-2) +(-3)(0). h. (-4) (-5) -(_2)(-l). i. (+2) (-3) -(+7) (-5). 4. Determine the negative of each of the following : a. 5. b. - 3. c. 0. d. 2x. e. 2/3. f. 2-3. g. 2a - 36. h. - (x - y). i. -[(a) (-6)]. j. Sx +2. k. x - 3. I. a + 0. 5. Find the reciprocal of each of the following: * a. ], b. 2/3. c. 3 +|- d. 2 4-i- e. 1.02. f. a +6. g. — • h. • * 3 a i. — ; — ' j. x - 2j. k. * x+p J# Jt r-0.1 2-0.3& T2 Introductory Topics Sec. 1-7 6« Prove each of the following equations by using rules for signs and for operations with fractions: a. a • ( - b) + a • ( - c) = - a • (6 + c). b. - (ac - ad) = a[d -f ( - c)]. c. - [6 - (a - c)] = (a - 6) - c. d. 6 /— - h c = a. e# . = . f. c [ a ~ ( a - 6/c)] = 6. b c - a c — a 1-8. ORDER RELATIONS FOR REAL NUMBERS We shall use the notation a > to express the fact that a is a positive number, and the notation a < to indicate that a is nega- tive. The symbol > means is greater than, and < means is less than. These symbols are called order symbols. Assume that a and b are any two given numbers. If a - b > 0, we shall write a > b, or b < a, and shall read "a is greater than b" or "6 is less than a." As can be easily seen, a > b means that a lies to the right of b on the real-number line. When a > b, that is, when a - 6 > 0, then 6 — a, which equals -(a - b) by (1-18), is nega- tive; and conversely. Hence, a>&(or6<a) if and only if b - a < 0. The student is familiar with the symbol = (for equality), which is used to indicate that two quantities are the same. Thus a = b means that the two symbols a and b represent the same mathe- matical object. For example, 6 = 3*2. If a and 6 are two distinct numbers on the scale, we say "a is different from b" or "a does not equal b" and we write symbolically a¥=b. The symbol ¥= means does not equal and is called the inequality symbol \ In general, the oblique line or vertical line through any symbol will form a new symbol which is the negation of the original one. Thus, a < b means "a is not less than b." In other words, a = 6 or a > b (by Property 1 below). For example, 5 < 3. Sometimes we shall find it convenient to combine the symbols < and = or > and =. We write ^ to mean is less than or equal to, and we write ^ to mean is greater than or equal to. We thus have order relations on pairs of real numbers, defined by either of the following equivalent statements : a > b (or b < a) if and only if a — b is positive; a > b (or b < a) if and only if b — a is negative. The system of real numbers is then said to be ordered by the rela- tion > (or the relation <) . Assertions of the type a < b or a > b are Sec. 1-8 Introductory Topics 13 called inequalities. The ordering of the real numbers has the fol- lowing properties. Property 1. For every pair of real numbers, a and 6, one and only one of the following relationships holds : a = by or a < b, or a > b. Proof of Property 1 : If a = b, the statement is certainly true. Saying that a ¥= b is equivalent to saying that a — b ¥= 0, so that a - b is either positive or negative. Thus, if a — b is positive, we have a > b. If, however, a — 6 is neither positive nor zero, then it is negative, and a < b. If two of the three possibilities occurred together, we should have, say, a = b and a > b, or a > b and a<b. Thus, a — b would be both zero and positive, or both positive and negative. Since no overlapping may occur among the three classes of numbers, we are thus led to a contradiction. Property 2. For any real numbers a, b, c, it is true that if a < b and b < c, then a < c. Proof of Property 2: If a< b and b < c, then both 6 — a and c — b are positive. Let us write c — a as (c - 6) + (b — a). We have assumed that the sum of two positive numbers is positive. Since c — b and b — a are positive by assumption, their sum, which is c — a, is also positive. Hence, c > a, or a < c. Property 3. For any real numbers a, b, c, it is true that if a > 6, then a + c > b + c. Proof of Property 3: By definition, a> b means that a — 6 is positive. But, by (1-17), (a + c) - (6 + c) = a + c + (-6) + (-c) = a - 6. It therefore follows that (a + c) - (b + e) is positive, and that a + c > b + c. Property 4. For any real numbers a, 6, c, it is true that if a > b and c > 0, then ac > 6c. Proo/ o/ Property 4: We have assumed that the product of two positive numbers is positive. Since both a - b and c are positive, it follows that their product is also positive. But (a— b)c = ac — be. Therefore, ac — 6c is positive, and ac > be. Property 5. For any real numbers a, b, c, it is true that if a > b and c < 0, then ac < be. 14 Introductory Topics Sec. 1-8 Proof of Property 5: We have noted that the product of two numbers with unlike signs is negative. Here a — b is positive, but c is negative. Since (a- b)c = ac — bc, and (a - 6) c is negative, it follows that ac — bc< 0, or that ac < be. According to Property 3, the order symbol in an inequality is not changed if the same number is added to or subtracted from both sides. It therefore follows that a term on one side of an inequality may be transposed to the other side with its sign changed. For example, if a — 6 > c, then a > c + b. According to Property 5, the order symbol in an inequality is reversed if both sides are multiplied or divided by the same nega- tive number. 1-9. ABSOLUTE VALUE As a consequence of the properties of the ordering of real num- bers, there can be associated with each number a certain non- negative number called its absolute value. For any real number a, we define the absolute value of a, denoted by \a\, as follows: | a | = a, if a ^ 0, and \a\ = — a, if a < 0. Thus, | 3 | = 3, since 3 > 0; also J — 3| = — ( — 3) =3, since - 3 < 0. 1-10. INEQUALITIES INVOLVING ABSOLUTE VALUES We shall now consider some inequalities involving absolute values. If we let the number x be represented by a point P on a number scale, then \x\ is the numerical distance between P and the origin. If we let a be a positive number, then \x\ < a means that the point P is less than a units from the origin; that is, x lies between -a and a. We can write this in the form -a < x and x < a, or more briefly in the form — a < x < a. Therefore, the statements \x\ < a and —a<x<a mean exactly the same thing. A more general inequality which often occurs is \x — b\<a, where a > 0. This is equivalent to —a < x — b < a. If 6 is added to each term, we may write 6 — a< x < b + a. Hence, the state- ments \x — b\<a and 6 — a<x <b + a mean exactly the same thing. For example, |#-3|<2 may be written -2<#-3<2 and means that the distance between x and 3 is less than 2. To solve this inequality for x, we add 3 to each term of the inequality, obtaining 1 < x < 5. The following illustrative examples may help to give a better understanding of the processes involved in the solution of the prob- lems in Exercise 1-2. Sec. 1-10 Introductory Topics 15 Example 1-5. Arrange the following numbers in increasing order: 2, - 3.5, 0, 7T, 3.14, |-5|. Solution: Since w is approximately 3.1416, the desired order is as follows: - 3.5, 0, 2, 3.14, 7T, |-5|. Example 1-6. Insert the proper inequality sign (order symbol) between the following numbers: - 2 and |-2 |. Solution: Since | - 2 ] =2, and since - 2 < 2, we have the inequality — 2 < |-2 |, or |-2 | > -2. Example 1-7. Find integers a and b such that a < \/2 < b. Solution: Since \/2 may be represented approximately by 1.414, the values a = 1 and 6=2 satisfy the inequalities. Thus, 1 < \/2 < 2. Any other pair of integers a and b such that a ^ 1 and 6^2 would also satisfy the inequalities. Example 1-8. Express the inequality | x j < 3 without using the absolute-value symbol. Solution: We know that the statements | x | < a and — a < x < a mean exactly the same thing. Here a is the positive number 3, and | x \ < 3 means that the point represented by x is less than 3 units from the origin; that is, x is between — 3 and 3. The inequality may be written — 3 < x < 3. Example 1-9. Explain the meaning of the inequality | x - 2 | < 1 and write it without using the absolute-value symbol. Solution: The inequality | x — b \ < a is equivalent to— a < x — b < a. Hence | x — 2 | < 1 may be written — 1 < x — 2 < 1. If we add 2 to each term of the inequalities, we obtain 1 < x < 3. EXERCISE 1-2 1. Arrange the numbers in each of the following sets in increasing order: a. - 3, 0, 4, - 2, 5. b. - 6, - 8, 2, 0, 1/2, - 3/4. c. - 2, 10, - 1, - 1/3, - 4. d. - 10, 9, 4, - 3, 3/8, - 6/5. e. 3, - 2, 1, V3, - 3/2. f . 1.4. 0, - 2, y/% | - 3 |. 2. Insert the proper order symbol between the two numbers in each of the following : *. 3 and 1/3. b. - 3 and | - 3 |. c. y/2 and 1.414. d. - 3 and - 2. f e. 22/7 and x. f. 1/8 and 1/6. 3. Examine each of the following inequalities, and determine whether or not it is true. a. - 5 > - 3. b. - 3 + 2 < 0. . c | - 3 | > - 3. d. t > 22/7. e. | - 2 | < | 2 |. f. | 3 - 7 | > | 5 - 2 |. 16 Introductory Topics See. 1-10 4. Find the value of each of the following: a.|+2|-|-2|. b.|-3| + |+3|. c|+4|-|-4|. d. | - 7 | + | - 5 | - | + 5 |. e. |12 - 4 | - | - 6 |. f. |5-3|+|3|-|2|. j. (-18) +3. h.|-9|+|4|. i. -s- 14. j. | 4 | -| -5|. 5. Express each of the following inequalities without using the absolute-value symbol: a. | x | < 1. b. 5 < 1. c. | x | ^ a. d. | 2x | < 4. e. | x - 1 | < 3. f. | x - 1/2 | < 3/2. 6. In each of the following, find a pair of integers, a and 6, such that the given inequalities are satisfied : a. a < 5 < b. b. a < - 3 < b. c. a < < b. d. a < 7T < 6. e. a < \/3 < 6. f . a < 1 1 - 2 | < 6 7. If a ^ 3, place the proper order symbol between a + 7 and 10. 8. If a ^ 5, what can be said about the value of 3a — 2? 1-11. POSITIVE INTEGRAL EXPONENTS If two or more equal quantities are multiplied by one another, the product of the equal factors is called a power of the repeated factor. Thus 5 2 , read "5 squared," means 5 • 5 ; 5 3 , read "5 cubed," means 5 • 5 • 5. In general, a n means the product of n factors each equal to a. We call a the base and n the exponent of the power. It follows from the associative law that a 2 • a 3 = (a • a) (a • a • a) =a,'a'<fa*a = a 5 = a 2+3 . Also, if a ** 0, a° a • a • a • a* a = a • a = a 2 = a 5 "" 3 . cr a • a • a These and similar results suggest the following laws of expo- nents. In them m and n are positive integers. The proofs of these laws are reserved for a later chapter. Law of Multiplication. To multiply two powers of the same base, add the exponents : (1-31) a m • a n = a m+n . Law of Division. To divide one power of a given base by another power of the same base, subtract the exponents : Q m (1-32) =j = or-*, if o * 0, m > n. Sec. 1-12 Introductory Topics 17 Law for a Power of a Power. To raise a power of a given base to a power, multiply the exponents : (1-33) (a m ) n = a mn . For example, (a 3 ) 2 = a 3 * a 3 = a 3 ' 2 = a 6 . Law for a Power of a Product. To obtain a power of a product, raise each factor of the product to the given power : (1-34) (a&)" = a n b n . Thus, (3a 3 ) 2 = 3 2 (a 3 ) 2 = 3 2 a 6 = 9a 6 . Law for a Power of a Quotient. To obtain a power of a quotient, raise the numerator and the denominator to the given power : (1-35) ©"=£' if6 *°- fa 2 \ 3 _ (a 2 ) 3 _ a 6 lhus,if&*0, (^) =^j = — , 1-12. ALGEBRAIC EXPRESSIONS An algebraic expression is formed by combining numbers by means of the fundamental operations of algebra. The distinct parts of the expression connected by plus and minus signs are called terms. The terms of the expression 3x 2 -5xy 2 +7z are 3x 2 y — 5xy 2 , and 72. Here the numbers 3, -5, and 7 are called numerical coffi- cients, or just coefficients; x 2 , xy 2 , and z are called the literal parts. An expression containing one or more terms is called a multi- nomial. A multinomial consisting of one term is a monomial. A binomial is a multinomial consisting of two terms, and a trinomial is a multinomial with three terms. A polynomial is a multinomial whose terms are of the form ax m y n z p • • • , where m, n, p, • • • are positive integers and a is a numerical coefficient, and where one or more of the factors x m , y», z p , • • • may be absent. Thus, 7, 5# 4 , and Sxy + 2 are polynomials, while x + - is not. y The degree of a term of a polynomial is the sum of all the expo- nents in its literal part. For example, the degree of Sx 2 is 2, the degree of — 5xy 2 is 3, and the degree of lz is 1, because the sums of the exponents are, respectively, 2, 3, and 1. The degree of a polynomial is the degree of its highest-degree term. Thus, in the trinomial 3x 2 -5xy 2 + 7z, the third-degree term, ~5xy 2 9 is its highest-degree termf Therefore, Zx 2 - 5xy 2 + 7z is a polynomial of the third degree. 1 8 Introductory Topics Sec. 1-12 By a polynomial in x of degree n we mean an expression of the form a x n + a\x n ~ l + • • • + a n , where the coefficients a , a u • • • , a n are numerical coefficients, a t* 0, and n is a positive integer. If n = 0, we agree that the poly- nomial reduces to a number a , which is not 0, and that the degree is zero. The number is regarded as a polynomial also, but as one having no degree. If the typical polynomial just given has degree n, the coefficient a is called the leading coefficient. If its leading coefficient is 1, a polynomial is called monic. Thus, x* — 2x 2 + bx + 7 is a monic poly- nomial of degree 3. 1-13. EQUATIONS AND IDENTITIES An equation is a statement of equality between two numbers or algebraic expressions. The two expressions are called members, or sides, of the equation. Equations are of two kinds, namely, conditional equations and identities. A conditional equation, or simply an equation, may be true only for certain values (possibly none at all) of the literal quantities appearing. An identity is true for all numerical values that can be substituted for the literal quantities. Illustrations of equations are 3x - 5 = x + 1 and x 2 - bx + 4 = 0. The first one is true only if x = 3, and the second is true only if x = 1 or x = 4. Illustrations of identities are 3(x - 2) = 3z - 6 and x 2 — 5x + 4 = (x — 1) (x — 4). Each of these equations is true for all values of x. 1-14. SYMBOLS OF GROUPING Parentheses ( ) and other symbols of grouping which have the same meaning as parentheses, namely, brackets [ ], braces { }, and the vinculum , are used to associate two or more terms which are to be combined to form a single quantity. The word "parentheses" Sec. 1-14 Introductory Topics 19 is often used to indicate any or all of these symbols of grouping. Removal of the symbols of grouping is accomplished by applying the laws of algebra, such as the laws of signs and the distributive law. The following examples illustrate the procedure. Example 1-10. Remove parentheses from - (2x — 3). Solution: The steps may be indicated as follows: - (2x -3) = (- 1) (2x -3) = (-l)(2s)+(-l)(-3) = - 2x + 3. Exampl e 1-11. Remove symbols of grouping from Sx — 2[5y + 3 (x — y)] — {2y - x - Sy} and collect terms. Solution: One way of obtaining the desi red resu lt follows: Sx - 2[6y + 3(x - y)] - [2y - x - Zy] = Sx - 2[5y + 3x - 3y] - \2y - x + 3y} = Sx -2[2y +3x] - {5y - x} = Sx - iy - 6x - by + % = dx - 9t/. The basic rules for enclosing a group of terms in parentheses may be stated as follows : To write a given expression in parentheses preceded by a plus sign, write the terms as they are given, enclose them in parentheses, and write + in front of the parentheses. Thus, a — b = + (a — b). To write a given expression in parentheses preceded by a minus sign, change the sign of each term, write the resulting terms in parentheses, and write — in front of the parentheses. Thus, a — 6 = — ( — a + 6) = — (6-a). The first rule is obvious, and the second follows from the rule of signs (1-18). Example 1-12. a) Enclose the last two terms of 2 + Sx - y within parentheses preceded by a plus sign. 6) Enclose these terms within parentheses preceded by a minus sign. Solution: a) Since the sign before the parentheses is to be +, we enclose 3a - y in its given form within the parentheses preceded by a plus sign. In this case the ex- pression becomes 2 4- (3x - y). 20 Introductory Topics Sec. 1-14 6) Since the sign before the parentheses is to be -, we change the sign of each term of Sx — y and enclose — 3# 4- y within the parentheses preceded by a minus sign. Then the expression becomes 2 — (- Zx + y). 1-15. ORDER OF FUNDAMENTAL OPERATIONS Parentheses and other symbols of grouping are useful in indicat- ing which operation is to be performed first. We have used them in this way from the outset. In order to avoid using them unneces- sarily, as has been already pointed out, the convention is adopted to perform all multiplications first and then the additions (or subtractions). If two or more of these symbols of grouping are used in the same expression, we usually (though not necessarily) remove the innermost pair of symbols first. Illustrations in which symbols of grouping are removed follow : a) 4 - (6/2) =4-3 = 1. b) (4 - 6)/2 = (- 2)/2 = - 1. c) 6 + [15/(3 • 5)] = 6 + (15/15) =6 + 1=7. d) 6 + (15/3) -5 = 6 + 5-5 = 6 + 25 = 31. operations. 1. 2«. 5. 10». 9. 7*. 13. (!)'• 17. !<*'>• 21. a 3 a 4 . 25. (a 2 6 3 ) 2 . 29. (a n b m ) 2 , 33. 2l. EXERCISE 1-3 q the group from 1 to 36, perform the indicated operation oi 2. 3*. 3. (- l) 3 . 4. 5 3 . 6. (-2)*. 10. (-4)'. 7. (-6) 3 . "• S) • 8. - 10 6 . »• (!)'• »(-!)• 16. 4(2°). 18. |(3«) • 19. a(5 8 ). 20. a 2 6(-2) 3 , 22. (a»)«. - 23. (a&) 4 . 24. a*a 7 . 26. a -a*- a 3 . 27. a 2 • a* • a 7 . 28. [(ab)*]*. 30. (a*& 2 )». a 6 31. V a 2 88.2!. a 3 «■<»■ *©• *©'• In each problem in the group from 37 to 48, remove symbols of grouping and simplify. 37. 3 - (6 - 2). 38. x + (y - *). 39. 4[a + (6 - a)]. 40. (a + 26) - (3a - b). 41. a - & + (2a - 36). 42. a - [3 + (2a - 4)]. 43. (a - 26) - (3a + 6). 44. 6s* - [Sx + (2y - x*)]. Sec. 1-16 Introductory Topics 21 45. ax 2 - 2bxy + [(by 2 - 2cx 2 ) - (axy + y 2 )]. 46. 3n6 - {4ac + [ab - 2ac + ab] - 3a&}. 47. 2a - [36 + 4c - {3a - b + a - 6 - (3a 4- 2c) } - 4c + a]. 48. - { - [ - (a - ft - c) - a + (6 - c)]}. In each problem from 49 to 60, enclose the last two terms in parentheses. First use a plus sign before the parentheses, and then use a minus sign. 49. a + b + c 50. a 2 - 2aft + b 2 . 51. a 2 - b 2 + c 2 . 52. x + ?/ - 1. 53. 2a + b - 3c. 54. 3z - 4y + 2z. 55. x 2 - y 2 - z 2 . 56. - x 2 - ?/ 2 - z 2 . 57. - a 3 ft + aft + ft 2 . 58. ax 2 - 2a:n/ + y 2 . 59. 2z - 3?/ - 4z. 60. - x 2 - x + 1. In each of the following problems, evaluate the given expression. 61. 16 - (6 - 2). 62. 10 - 6 - 2. 63. (- 3) (-4) - (4) (- 2). 64. 4 • (6 - 7). 65. 4 • 6 - 7. 66. (4 • 6) - 7. 67. (3 • 3) - (4 • 2). 68. 3 • 3 - 4 • 2. 69. 3 • (3 - 4) • 2. 70. 3 • 3 - (4 • 2). 71. (8/2) + 4. 72. 8/(2 + 4). 73. 8 + (4/2). 74. (8 + 4)/2. 75. 8 + 4(1/2) + 3. 76. (8 + 4)/ (2 + 3). 77. 8 + (4/2) + 3. 78. 8 + ((4/2) + 3). "«• »(S?^/»)->- "•l^/c-"- -[f^-OA 1-16. ADDITION AND SUBTRACTION OF ALGEBRAIC EXPRESSIONS Terms, such as 2x 2 y 3 and 5x 2 y s , which have the same literal parts, are called similar or like terms and may be added or subtracted by adding or subtracting their coefficients. To illustrate, let us con- sider an example. Example 1-13. Add 5xy 2 , 7x 2 y, - 2xy 2 , - 9x 2 y, 4x 2 y*. Solution: Collecting like terms and adding coefficients, we have (5 - 2)xy 2 + (7 - 9)x 2 y + 4x 2 t/ 3 = 3xy 2 - 2x 2 y + 4x*y*. The procedure used in the solution of Example 1-13 follows at once from the distributive law. For example, the sum of the terms 5xy 2 and -2xy 2 is obtained as (5 - 2)xy 2 or Sxy 2 . This leads at once to the rule for the addition (or subtraction) of algebraic expressions. In practice, we usually arrange like terms in vertical columns, and then we find the sum of each column by prefixing the sum of the numerical coefficients in the column. The procedure may be made clearer by means of the following examples. 22 Introductory Topics Sec. 1-16 Example 1-14. Add 2x 2 - Sxy + z,x 2 - 5z, 2xy + 3z. Solution: 2x 2 - Zxy -f z x 2 - 5« 2xy + 3z 3a; 2 — xy — z. Example 1-15. Subtract 2a 2 - 36 + c 2 from 3a 2 - c 2 . Solution: 3a 2 — c 2 2a 2 - 36 + c 2 a 2 + 36 - 2c 2 . 1-17. MULTIPLICATION OF ALGEBRAIC EXPRESSIONS With the help of the distributive law for multiplication, the product of two algebraic expressions is found by multiplying each term of one by each term of the other and combining like terms. Example 1-16. Multiply 3x 2 - 2xy + y 2 by 2x - 3y. Solution: 3x 2 - 2xy + y 2 2x - Sy 6x* _ ± X 2 y + 2xy 2 - 9x 2 y + 6xy 2 - -3^ 6z 3 - lSx 2 y + &rz/ 2 -32/3. 1-18. SPECIAL PRODUCTS The following typical forms of multiplication occur so frequently that we should learn to recognize them quickly and to obtain the products without resorting to the general process of multiplication. They should be learned thoroughly. (1-5) a(b + c) = ab + ac. (1-36) (a + b) (a - 6) = a 2 - 6 2 . (1-37) (a + b) (a 2 - ab + b 2 ) = a 3 + 6 3 . (1-38) (a - 6) (a 2 + ab + b 2 ) = a 3 - 6 3 . (1-39) (a + 6) 2 = a 2 + 2a6 + fe 2 . (1-40) (a - 6) 2 = a 2 - 2a& + 6 2 . (1-41) (ax + by) (ex + dy) = acz 2 + (ad + bc)xy + bdy 2 . See. 1-19 Introductory Topics 23 1-19. DIVISION OF ALGEBRAIC EXPRESSIONS To divide a polynomial by a monomial, divide each term of the polynomial by the monomial and add the results. This rule follows immediately from the rule for fractions expressed by (1-25). Example 1-17. Divide 6a 2 x 2 - 12a% - 30a«x* by 15a 3 x 2 . 6a 2 x 2 - 12a 4 x - 30a 6 z 5 6a 2 z 2 12a 4 z 30a 6 z 5 Solution: 15a*x 2 15a 3 x 2 15a 3 # 2 15a 3 x 2 = f _ |a _ 2a3x3 _ ba ox To divide a polynomial (the dividend) by a polynomial (the divisor) , arrange both according to descending or ascending powers of some common literal quantity. Then proceed as follows: Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient. Multiply the entire divisor by the first term of the quotient, and subtract this product from the dividend. Use the remainder found by this process as a new dividend, and repeat the process. Continue the work until you obtain a remainder that is of lower degree in the common literal quantity than the divisor. Example 1-18. Divide 6a; 3 - 5x 2 + 3x + 1 by 2x - 1. Solution: 2s - 1 | 6x 3 - 5x 2 + 3x + 1 1 3x 2 - x + 1 6s 2 - 3s 2 - 2x 2 + 3x -2x 2 +x 2x + 1 2x - 1 2. The division can be checked by finding the product of (2x - 1) and (3x 2 - x -f 1) and adding 2, proving that 6x 3 - bx 2 + 3x + 1 = (2x - 1) (3z 2 - x + 1) + 2. Any problem in division, in general, may be checked by means df the relationship ? dividend = (divisor) (quotient) + remainder. This equation is an identity ; that is, it is true for all values of the literal quantities. Indeed, this equation supplies the underlying meaning of the process of division. 24 Introductory Topics Sec. 1-19 EXERCISE 1-4 In each problem in the group from 1 to 12, add the given expressions which are separated by commas. 1. - xy, Sxy. 2. ix 2 y 2 , - 2xy. 3. Sx 2 y 2 , - 2x 2 y 2 . 4. 6a +116, - 3a + 26. 5. 4a 2 - 2b 2 , a 2 - 46 2 . 6. 3a - 2b -f- 4c, 4a + 36 - 6c. 7. — 3a + b — c, — a — & 4- c. 8. 3x -b 2?/ - z, x + ?/ - Sz, 4x - 3y -f 2z. 9. x 2 - 2xy + ?/ 2 j 4x?/, - t/ 2 . 10. 3x 2 - 4xy 2 + 2?/ 3 , 4x 2 y - 2x 2 - y 3 , 4xy 2 - 2?/ 3 . 11. 2x 3 - 3x -f 1, x 2 4 2x - 3, 2x 2 - x 3 + 4 - 2x. 12. 4ax 4- 3bxy - Ax 2 , 2bx - bx 2 4 2axy, Sx - 2y. In each problem from 13 to 24, subtract the second expression from the first. 13. — xy, Sxy. 14. 4x 2 ?/ 2 , — 2xy. 15. Sx 2 y 2 , - 2x 2 y 2 . 16. 6a + 116, - 3a 4- 26. 17. 4a 2 - 26 2 , a 2 - 46 2 . 18. 3a - 26 4 4c, 4a + 36 - 6c. 19. - 3a + 6 - c, - a - 6 + c. 20. x 3 + x 2 4- x 4 1, x 3 - x 2 - x -1. 21. x 2 -f 2xy 4 2/ 2 , 4x?/. 22. 3 - 2x 4 x 2 - x s + 3x 4 - 2x 5 , 3x 5 4 4x 4 - x 3 -f- 2a: 2 - 2x 4 1 . 2 12 2 2 17 23. -a; 3 - -a; 2 -4- ^xy - -y 2 , -x 3 4 jxy - ^/ 2 ' 24. -xy 2 - =x 2 y 4- -xy 2 - ^y 2 > ^x 2 y - -xy 4- ^/ 2 • In each problem from 25 to 60, perform the indicated operations. 25. (6x)(-3?/). 26. (4x)(-2). 27. (5x 2 y) (- 2xy). 28. (3a) (26 2 ) (- 4c). 29. (- 3a6) (46c) (- 2a 3 ). 30. (x - y)2a. 31. (3x + 2y) (- 2x). 32. (6 - 3a) (2). 33. (4a5|/ 4 2y 2 ) (3xy). 34. (4a 2 + 66 2 ) (Sab). 35. (2x + Sy) (x - y). 36. (3a - 6) (a + 26). 37. 4x 2 (x 2 - 2xy - y 2 ). 38. Sxy(2x 2 y + Sxy - 4y 2 ). 39. (x 2 + xy - y 2 ) (x - y). 40. (x 2 - y 2 ) (x + y) (x - y). 41. 4x/(- 2). 42. 3x 2 y/2xy. 43. 4x 2 y 3 z/2xy 2 *. 44. 20x 6 y 4 2yi0x 6 yz 3 . 45. 4a6/(- 4a6). 46. (2xy) 2 /2xy 2 . '47. (2xy) 2 /2(xy) 2 . 48. Or 2 -2xy)/(-x). 49. (3x?/ 2 - 6xy + 9y 2 )/(- 3y). 50. (a: 2 ?/ 3 - 3x 2 y 2 + 4x 3 y 2 )/x 2 y 2 . 51. (x 2 + 6x + 5)/(x + 1). 52. (9x 2 - 6x + l)/(3x - 1). 53. (x 2 - y 2 )/(x - y). > 54. (x - y) 2 /(x - #)• 55. (x 4 - y 4 )/(x - y). 56. (x 3 4 3x 2 4 Sx + l)/(x + l) 2 . 57. (x 3 + 2/ 3 )/(x 4- j/). 58. (z + y)*/(x + y). 59. (x 3 - y 3 )/(x - y). 60. (z 4 + x 3 + 3x 2 + 2x + l)/(x 2 + 2). 61. Divide x 2 - y 2 - (x - y) 2 - (x - y) (x y) by x - y. 62. Divide (a + 6) 2 + 6(a 4- b) + 5 by a -f 6 + 5. 63. Divide x 4 + 4x 3 4- 6x 2 4- 4x + 1 first by x 4- 1 and then by x 2 + 2x 4- 1. 64. Multiply x 2 - 2x - 3 by x + 4, and divide the result by x 4- 1. Sec. 1-20 fnfrocfucfory Topics 25 65. Divide x 5 + x 4 + 3x 3 - 2# 2 - 3 by x - 1 and add the quotient to the excess of 3z 4 4- 2x 2 - 9x + 7 over 2z 4 + 2x 3 + 7x 2 + 3z + 4. 66. a) Under what conditions will ( - #) n be positive? 6) When will it be negative? Assume first that x is positive and then that x is negative. 67. For what values of n will (— a) n be equal to — a n ? 1-20. FACTORING Factoring a quantity is the process of finding quantities which, when multiplied together, yield the given quantity. When a quan- tity A is expressed as a product B • C, B and C are called factors or divisors of A and are said to divide A. Also, A is called a multiple of each of Z? and C. These concepts are applicable to numbers or algebraic expressions generally, but are most useful when restricted to apply to integers or to polynomials. Such restriction will be adhered to in this book. Thus, when an integer is to be factored, the factors sought are to be integers. And when a polynomial is to be factored, the desired factors are to be polynomials. Let us first review the fundamentals of factoring integers (posi- tive, negative, or zero) . First, it is clear that every integer n may be expressed as 1 • n or (-l)-(-w). Such factorizations are called trivial. If an integer n, other than 4-1 or — 1, has no factorizations other than trivial ones, then n is called a prime (number). An integer having a non-trivial factorization is called composite. Examples of prime integers are 3, 7, and —11; examples of com- posite integers are 6 and —40. Let n be a composite integer. Then a non-trivial factorization m 'p exists in which \m\ and \p\ are less than \n\ and greater than 1. If both m and p are primes, then n is expressed as a product of primes. If not, at least one of m and p, say p, is composite, and so p = r • s. Hence, n~ m • r • s. The process begun may be continued if any one of m, r, and s is composite, and additional factors may be found, until the process cannot be continued further, in which case only prime factors are obtained. We may conclude that the factoring process must terminate, since at any stage the new fac- tors introduced are numerically less than their product. The fundamental theorem of arithmetic guarantees that every com- posite integer is a product of primes, which &re unique except for their signs or the order in which they are written. For example, successive factoring of 156 gives 156 = 39 • 4 = 13 • 3 • 4 = 13 • 3 • 2 • 2, and the final factors 13, 3, 2, 2 are primes. Another valid factor- ization of 156 into primes is as follows : 150 = 2-3- (- 13) -(-2). 26 Introductory Topics Sec. 1-20 Let us turn now to polynomials with real coefficients. These may be polynomials in x, such as x 2 — 1 ; or polynomials in x and y, such as x 2 + Sxy + 2y 2 ; or, in fact, polynomials in any number of literal quantities. Every polynomial F has a factorization of the form F = l(aF), for every non-zero number a. And, of course, the factors 1/a and aF are themselves polynomials. Such factorizations are called trivial. If a polynomial F has no factorizations other than trivial ones, then F is called a prime polynomial, or an irreducible poly- nomial. A polynomial having a non-trivial factorization is called composite or reducible. Examples of prime polynomials are Sx + 2, 2x + 2y y and x 2 + 3. Every polynomial in x of the first degree may be shown to be prime; and certain polynomials of higher (even) degree in x also are prime. A development of criteria for primeness of polynomials lies beyond the scope of this book. Examples of composite polynomials are x 2 — 4, xy + y 2 and xz + yz + xu + yu, because each of these has a non-trivial factor- ization. Thus, we have x 2 - 4 = (x + 2) (x - 2), xy + ij 2 = (x + y)y, xz + yz + xu + yu = (x + y) (z + u). Let / be a composite polynomial in x with real coefficients, not all of which are zero. Then it can be shown that there exist poly- nomials g and h, the degree of each of which is less than that of /, such that f = g-k. If g and h are primes, then / is a product of primes. If not, we may proceed to factor (non-trivially) one or both of g and h, and we can continue this process until primes are obtained. The process must terminate eventually, since each non-trivial factorization leads to polynomials of lower degree. The argument just presented applies only to polynomials in one literal quantity x. However, the principle may be extended to apply to polynomials in any number of literal quantities x, y, z, • • • . Thus, / = Pip 2 --'Pn, where p lf p 2 , • • • , p n are prime polynomials. The problem of carry- Sec. 1-21 Introductory Topics 27 ing out actual factorizations for certain types of polynomials is considered in the next section. A special class of polynomials deserves particular attention. This is the class that consists of polynomials in which all the numerical coefficients are integers. It is possible to prove a factor- ization theorem such as that just stated, but yielding factors which are polynomials having only integral coefficients. When the general theorem can yield prime factors all of which have integral coeffi- cients, the two theorems give the same result. Otherwise, they will give different results. For example, both theorems applied to x 2 - 4 yield the prime factorization (x 4-2) (x -2). The polynomial Sx 2 - 4 has no non- trivial factors with integral coefficients. However, when other real coefficients are allowed, we have 3x 2 - 4 = (y/3x + 2) (y/Zx - 2), in which the coefficient \/S is a perfectly acceptable real number. Again, when factors with integral coefficients are desired in such a case as 36# + 24ty, we shall agree to remove and factor common numerical factors, to obtain 36x + 24y = 3 • 2 • 2 • (3x + 2y). Here the prime factors are the numerical primes 3, 2, 2 and the prime polynomial 3x + 2y. In what follows, whenever a polynomial has only integral coeffi- cients, we agree to restrict ourselves to factors of the same kind. In similar fashion, when the given polynomial has only rational coefficients, we shall search for factors having only rational coefficients. 1-21. IMPORTANT TYPE FORMS FOR FACTORING The equations in Section 1-18 applied "in reverse" are formulas for factoring. Success in factoring a polynomial therefore depends on ability to recognize the polynomial as being a particular type of product and as having factors of a definite form. Verify the follow- ing type forms by carrying out the indicated multiplications and learn each form. Type 1: Common Monomial Factor, From (1-5) we have (1-42) ab + ac = a(b + c). Example 1-19. Factor 4x 2 y - 6xy 2 . Solution: ±x 2 y - Qxy 2 = 2xy(2x - Zy). 28 Introductory Topics Sec. 1-21 Instructions for actually removing the monomial factor in Example 1-19 may be put this way : Write the common factor, 2xy, and in parentheses following 2xy write the algebraic sum of the quotients obtained by dividing successively every term of 4x 2 y - 6xy 2 by 2xy, in accordance with the distributive law. Type 2: Difference of Two Squares. From (1-36) we have (1-43) a 2 - b 2 = (a + b) (a - b). Example 1-20. Factor 9x 2 - 25s/ 2 . Solution: 9x 2 - 2hy 2 = (3x + by) (3x - by). Type 3: Sum and Difference of Two Cubes. From (1-37) and (1-38) we have (1-44) a 3 + 6 3 = (a + b) (a 2 - ab + b 2 ) and (1-45) a 3 - 6 3 = (a - b) (a 2 + ab + b 2 ). Example 1-21. Factor &r 3 - 27y*. Solution: 8x* - 27 y* = (2x - Zy) [(2a;) 2 + (2s) (Zy) + (3?/) 2 ] = (2x - Sy) (4z 2 -f 6x2/ + 9y 2 ). Type 4: Perfect-Square Trinomials. From (1-39) and (1-40), (1-46) a 2 + 2ab + b 2 = (a + 6) 2 and (1-47) a 2 - 2a6 + b 2 = (a - 6) 2 . Example 1-22. Factor 4x 2 + 12xy 2 + 9y 4 . Solution: Ax 2 + 12;n/ 2 -f 9?/ 4 = (2a;) 2 4- 2(2s) (3z/ 2 ) + (3y 2 ) 2 = (2* + Sy 2 ) 2 . Type 5: General Trinomial. Factorization of the general tri- nomial may be indicated as follows : (1-48) Ax 2 + Bxy + Cy 2 = (ax + by) (ex + dy). If the two factors {ax + by) and (c# + dy) are multiplied together, the product is found to be acx 2 + (be + ad)xy + bdy 2 . By comparing this product with the trinomial Ax 2 + Bxy + Cy 2 , we note that it is necessary to find four numbers a, b, c, d, such that ac = A, be + ad = B } and 6d = C. Sec. 1-21 Introductory Topics 29 The following example illustrates a trial-and-error procedure, which often involves several steps of inspection and testing by multiplication ; yet it is a method commonly employed in practical work and is recommended here. Example 1-23. Factor 6x 2 + llxy - 10y 2 . Solution: Here we wish to find a, 6, c, and d which satisfy the identity (ax + by) (ex + dy) = acx 2 + (ad + bc)xy + bdy 2 = 6x 2 + llxy - 10?/ 2 . Since ac = 6 and bd = — 10, obviously a and c have like signs, and b and d have opposite signs. Possible values for a and c are ± 1, ± 2, =fc 3, and =fc 6. Possible values for 6 and d are db 1, =fc 2, ± 5, and =fc 10. By trial and error we find the correct selection to be a = 2, c = 3, b = 5, and d = - 2. This selection meets the requirement because (2* + 5y) (3x - 2y) = 6z 2 + llxy — I0y 2 . Type 6: Factoring by Grouping. An expression which does not fall directly into one of the given type forms can sometimes be reduced to one of these forms by a suitable grouping of terms. The following examples illustrate the procedure. Example 1-24. Factor Sax - 5bx + Gay - 1(%. Solution: First, group within parentheses the terms having a common factor. Thus, Sax - 5bx + Say - IQby = (Sax - 5bx) + (6a?/ - 1(%). Then, in each group, factor out the common quantity. In this case, (Sax - 5bx) + (Sax - 1(%) = x(Sa - 5b) + 2y(Sa - 56). Finally, factor out the quantity common to the terms obtained. (This quantity will often be a multinomial factor.) The result is x(Sa - 56) + 2y(Sa - 56) = (3a - 56) (x 4- 2y). An alternate method of grouping would give us the following results : First, 3ax - 5bx + Say - lOby = (Sax + Say) + (- 56s - 1(%). Then, (Sax 4- Say) 4- (- 56s - 1(%) = 3a(x + 2y) - 56(z + 2y). Finally, 3a(z + 2^/) - 5b(x + 2y) = (x + 2y) (3a - 56). Example 1-25. Factor z 2 - ?y 2 4- 62/0 - 9* 2 . Solution: By grouping the last three terms, we may rewrite the given expression as the difference of two; squares. Thus, x 2 - y 2 + Syz - 9z 2 = x 2 - (y 2 - 6^ + 9s 2 ) = z 2 - (2/ - 3*) 2 = [* - (y - 3*)1 [* + (y - 3«)1 = (z - 2/ 4- Sz) (x + y ~ 3*). 30 Introductory Topics Sec. 1-21 Example 1-26. Factor 4s 4 + 3x V + s/ 8 . Solution: By adding and subtracting »V, we may rewrite the given expression as the difference of two squares. Thus, 4 X 4 + 3^4 + y s = 4s 4 + 4s 2 ?/ 4 + </ 8 - * 2 2/ 4 = (2s 2 + 2/ 4 ) 2 - xW = [(2s 2 + t/ 4 ) - xy*] [(2s 2 + 2/ 4 ) + xy*] = (2s 2 4- y 4 - Z2/ 2 ) (2s 2 + 2/ 4 + si/ 2 ). EXERCISE 1-5 Factor each of the following: 1. 3s + 62/. 2. 6s 2 + 4s2/. 3. 4s + 14. 4. 2a - db. 5. 3as + 2a. 6. 4as 2 - 2/a 2 s 3 . 7 - as -f 2cs - s 2 . 8. 2a* - 360* + 6c* 3 . 9. ax - 2ay + 3a*. lol 3a6 + 9ac - 66c. 11. W + 3y» - at/ 2 . 12. s 3 - 5s 2 2/ + 6s?/. 13 a; 2 - 16. 14. y 2 - 64. 15. 4s 2 - 9. 16. 16s 4 - 25y*. 17. 49s 2 - 121. 18. 25a 2 - 166 2 . 19. 9y* - a 2 . 22. 4a 2 s 4 - 81. 25. 49s 2 ?/ 2 - 144a 2 6 2 . 26. a 2 - ±- 27. 36s - s* 28. 4s 2 31. • 20. 1 - s 2 ?/ 2 . 21. a 2 s 4 - 9a 4 s 2 . 23. 0.01 - & 2 . 24. - a^x 2 - ax*. - 16s 4 . 29. 2s 3 + 16. 30. 3s 3 - 81?/ 3 32. a» - 2166». 33. 125pV + r 16 . 34 a' + 6". 35. x« - 2/> 2 . 36. 2»V - 128^. 37*. Six" - 3(22/ + &)». 38, (a + 26)3 +.(3 C ^4d)'.__3?^( ? _ i ^- (2- ^) 3 - "> ^ „.,' " '-"41. 216«» -l/< 42 - * 2 - "* + 3b - ~^^ 12x3 + *.. 44. 8x2 + 2x - 15. 45. xV - l&W + 81. - 6 47. x 4 - 8x^ + 15. 48. x* - 11* + 30. £T-~2x - 3. . 50. 2x* - 3x + 1. 51. x* - 4x</ + 4tf*. „. x" - 2* 2 + 1. 53. x* - x - 12. • 54. x* - 3x - 10. 55! x* + 3*2 + 2. 56. *• + 4x3 + 4. 57. 1 + 30x3 + 225 xe. 58 x< - 10x= + 9. -5Q. 5x3 + 10x3 _ 40x. 60. a' + ba - 24. 6l! 2x2 _ nx _ 6. 62. x2 + 16x + 64. 63. 2x2 + 5x - 3 64. 18x2 + i5x - 25. 65. 6x2 _ 3 7x + 6. 66. 4x2 + 32 x + 15. 67 x2 - 1 2x + 0.36. 68. 4<te*tf + 35xyz - 1&*. 69. ax + 3x + 2ay + Qy. 70. x3 + 3x2 _ 7 x - 21. 71. 8x» _ IV - lto + 15. 72. ax + 26* - ex + 3ay + 66*/ - 3cy. 73. 2ax + 3y - xz + 6a + xy - 3z. 74. xa6 - xyz - 2a6j/ + 2y 2 z. 75. x 3 + x 2 - 3x - 3. 1-22. GREATEST COMMON DIVISOR A common divisor of several polynomials (or integers) is a poly- nomial (or integer) which divides each of them. For example, 2 and 3 are common divisors of 12 and 18. Also, x and x + y are common divisors of x* (z 2 - y") and x 3 (x 2 + 2xy + y>). A greatest common divisor (G.C.D.), also known as a highest common factor (H.C.F.), of two or more polynomials (or integers) Sec. 1-22 Introductory Topics 31 is a polynomial (or integer) with the following two properties: It is a common divisor of the given polynomials (or integers) ; also it is a multiple of every other common divisor of the given polynomials (or integers). It follows that, for integers, a G.C.D. is a common divisor of greatest absolute value. It also follows that, for polynomials, a G.C.D. is a common divisor of highest degree. For example, a G.C.D. of 12 and 18 is +6 or -6, since ±1, ±2, ±3, and ±6 are the only common divisors, and 6 and —6 are those of maximum absolute value. This example indicates that a set of non-zero integers will have two greatest common divisors, d and — d, one being positive and the other negative. For polynomials with real coefficients, if d is a G.C.D., then a • d is also a G.C.D. for every real number a not equal to 0. It follows that infinitely many greatest common divisors exist. However, for polynomials with integral coefficients, a G.C.D k should be a poly- nomial of the same type. Example 1-27, which follows, illustrates the fact that in this case a G.C.D. is uniquely determined except for sign. Since a G.C.D. of given polynomials divides all of them, it must contain as a factor each of the distinct prime factors occurring as a common factor of all the given polynomials. Since, however, a G.C.D. must divide any common factor of the polynomials, it must contain each of the distinct common primes raised to the highest common power. Thus, a G.C.D. of x 3 (x + y) 2 and x* (x — y) (x + y) must contain the prime factor x to the third power, that is, to the highest common power. Similarly, it rftust contain (x + y) to the first power. Thus, x*(x + y) is a G.C.D., since no further common prime factors occur. When no common prime factors occur, a G.C.D. is 1. Example 1-27. Find a G.C.D. of 3x*y*(x* - 4s/ 2 ) and Sxy 2 (x 2 - 4xy -f 4z/ 2 ). Solution: We shall begin by writing each of the expressions as the product of its prime factors, as follows: Zxhjs(x* - ±y*) = (3) (x) (x) (y) (y) (y) (x -f 2y) (x - 2y), and 6xy*(x* - ixy + ±y 2 ) = (2) (3) (x) (y) (y) (x - 2y) (x - 2y). The different prime factors are 2, 3, x, y, (x -f 2y), and (x - 2y), of which only 3, x } y } and x - 2y are common to both polynomials. We now form the product of these common factors, using for each the maximum common power. Hence, a G.C.D. is 3xy 2 (x - 2y). 32 Introductory Topics Sec. 1-23 1-23. LEAST COMMON MULTIPLE A common multiple of two or more polynomials (or integers) is one containing each of the given ones as a factor. Thus, 36 is a common multiple of 6 and 9, and x 2 — y 2 is a common multiple of x — y and x + y. A least common multiple (L.C.M.) of two or more polynomials (or integers) is a polynomial (or integer) with the following properties: It is a common multiple of the given polynomials (or integers) ; also it is a divisor of every other common multiple of the given polynomials (or integers). It follows that, for integers, an L.C.M. is a common multiple of least absolute value. It also follows that, for polynomials, an L.C.M. is a common multiple of lowest degree. For example, an L.C.M. of 6 and -8 is 24 or -24, since the only common multiples are ±24, ±48, ±72, • • • , and 24 and -24 are those of minimum absolute value. This example indicates that a set of non-zero integers will have two least common multiples, m and — m, one being positive and the other negative. For polynomials with real coefficients, if m is an L.C.M., then a • m is also an L.C.M. for every real number a not equal to 0. It follows that infinitely many least common multiples exist. How- ever, for polynomials with integral coefficients, an L.C.M. should be a polynomial of the same type. Example 1-28, which follows, illus- trates the fact that in this case an L.C.M. is uniquely determined except for sign. Since an L.C.M. of given polynomials is a multiple of all of them, it must contain as a factor each of the distinct prime factors occur- ring as a factor of any one of the given polynomials. Since, how- ever, an L.C.M. must be a multiple of any common multiple of the given polynomials, it must contain each of the various distinct primes to the highest power occurring anywhere. For example, an L.C.M. of x s (x + y) 2 and x*(x - y) (x 4- y) must contain the vari- ous distinct prime factors, which are x, x + y, and x - y. For x, the highest power occurring anywhere is the fifth power ; f or x + y, the highest power is the second ; for x — y, the highest power is the first. An L.C.M. is therefore x 5 (x + y) 2 (x-y). Example 1-28. Find an L.C.M. of 4x2 - ± Xy § X 2 _ 6, and 9x 2 - 18x + 9. Solution: We shall first rewrite each of the expressions in factored form. Thus, 4x2 - 4x = (2) (2) (x) (x — 1) = 2 8 x(x - 1), 6x2 - 6 = (2) (3) (x + 1) (x - 1), and 9x2 _ 18x + 9 - ( 3 ) ( 3 ) ( x - i) (s - i) = 32(x - 1)2. Sec. 1-24 Introductory Topics 33 The distinct prime factors are 2, 3, x, x + 1, and x - 1. The greatest powers for these are 2, 2, 1, 1, and 2, respectively. An L.C.M. is therefore 2 2 • 3 2 • x • (x + 1) • (3 - l) 2 . EXERCISE 1-6 In each problem from 1 to 12, find a G.C.D. of the given expressions. 1. 4, 14, 36. 2. 9, 21, 33. 3. 4, 7, 39. 4. x + y, x 2 - 2/ 2 . 5. x 3 - 2/ 3 , x - y. 6. 3a6, 12a 3 6, 6a 3 6 2 . 7. 9x 3 */ 2 , 15x 4 */ 2 , 21x 6 i/. 8. x - 3, x 2 - 9, (x - 3) 2 . 9. ±x*y% 8xij 7 z*, Ux s y*z 2 . 10. x + 2, x 2 - 4, x 3 + 2x 2 - 4x - 8. 11. x 3 - x 2 - 42x, x 4 - 49x 2 , x 2 - 36. 12. x 4 + 2x 3 - 3x 2 , 2x 6 - 5x 4 + 3x 3 , x 3 + 3x 2 - x - 3. In each problem from 13 to 22, find an L.C.M. of the given expressions. 13. 6, 8, 12. 14. 8, 45, 54. 15. xy, 6x2, 8yz. 16. 4x 2 , 5x 4 , 20x. 17. 4x 2 i/ 4 z 5 , 9x 6 i/ 2 2 3 > 6x 4 yz*. 18. 2a + 4, a - 3, a 2 - 9. 19. x - 2, x + 2, x 2 - 4. 20. 2x + 8, 3x - 6, x 2 + 2x - 8. 21. (x 2 - 49) (x 3 - 8), (x - 7) (x + 7) (x - 2) (x 2 - 4), (x - 3) (x - 2). 22. 2x 4 - 22/ 4 , 6x 2 + \2xy + 6y 2 , 9x 3 + 9yK 1-24. REDUCTION OF FRACTIONS From (1-24) under operations with fractions, it follows that a fraction a/b, in which b ¥=0 t is not changed if both the numera- tor and the denominator are multiplied or divided by the same quantity, provided that the quantity is not zero. That is, if k ¥= 0, n nn\ a - a / k (1 ~ 50) b~W For example, i _ 2^ _ 2 H * - ^ = ? 2 "" 2 • 2 " 4 ' 6 " 6/2 3 ' The fundamental principles of (1-49) and (1-50) are applied in reducing a fraction to lowest terms and in changing two or more fractions with different denominators into equivalent fractions with a common denominator. The reduction of a fraction to lowest terms, that is, to a form in which all common non-zero factors are removed from both the numerator and the denominator, is accomplished as follows: First, factor both the numerator and the denominator into prime factors. Then, divide both the numerator and the denominator by all their common factors. 34 Infroductory Topics Sec. 1-24 It should be noted that this reduction can also be accomplished by dividing the numerator and the denominator by their highest common factor. The following examples will illustrate the reduction of fractions to lowest terms. 30 Example 1-29. Reduce jt to lowest terms. Solution: Factoring the numerator and the denominator and dividing both by their common factors, we have 30 _ 2 ; 3 • 5 5 42 2 • 3 • 7 7 ' The common factors of the numerator and the denominator are 2 and 3. Example 1-30. Reduce tr-f- to lowest terms. 2x 2 y Solution: Factoring into prime factors and dividing out common factors, we have 2 • 3 ; x * y • y _ 3y # 2 • x • x • y x In this fraction the common factors are 2, x, and y. Example 1-31. Reduce -z-z ^ ^-r-r to lowest terms. 6x 2 - 3xy - ISy 2 Solution: We have 6s 3 - 24xy 2 2 * 3x(x - 2y) (x + 2y) = 2x(x -f 2y) , 6z 2 - 3sy ~ ISy 2 ~ 3(s - 2y) {2x + 3y) "" 2x 4- 3y It is important to note that in a fraction of the type a ~T - > where the numerator and the denominator have a common term, any attempt to simplify the fraction by cancelling out this common term can lead only to an absurdity. For, quite obviously, a y .. . a -j- x does not equal either "T - or - unless a = 1 or 0. For example, if 1 -f- X X o I o we attempt one of these simplifications with the fraction > 5 4 5 3 ^" we reach the obvious contradiction - = - or - = - • To avoid this 6 5 6 4 common error, it is important to remember that only common factors, not common terms, may be cancelled. One should make certain that the common quantity is a factor of the entire numera- tor and of the entire denominator. 1-25. SIGNS ASSOCIATED WITH FRACTIONS It follows from (1-49) that we can multiply both the numerator and the denominator of a fraction by -1 without changing the Sec. 1-25 Introductory Topics 35 value of the fraction. However, if just one of these is multiplied by — 1, then the sign of the fraction must be changed in order to keep its value unaltered. Thus, in effect, a minus sign before a fraction can be moved to either the numerator or the denominator without altering the value of the fraction. We have, as previously- stated in (1-22), n __ — n _ n , d ~~ d ~~ — d and i-ii i x - y x - y - (x - y) y - x We see, then, that any two of the three signs associated with a fraction, namely, the signs of the numerator, the denominator, and the fraction, can be changed without changing the value of the fraction. In general, the rules for changing signs in a fraction are as follows: Changing the signs in an even number of factors in the numera- tor or in the denominator, or in both, does not change the sign of the fraction. Changing the signs in an odd number of factors in the numerator or in the denominator, or in both, does change the sign of the fraction. 2x — 2x ( ) Example 1-32. Find the missing quantities in — = - -, r = -^-r • ,-j 7 . . ZlX ZlX £X Solution: — = — = ~ • o o — o Example 1-33. Change the fraction — — to an equivalent fraction with denominator x — y. Solution: Since x - y = - (y - x), we make the following changes in signs: a ___ - a __ - a V - x ~~-(y-x)~x-y' a a a y - x - (y - x) x - y EXERCISE 1-7 1. Find the missing quantity in each of the following equalities: a * 7 ~ 21 15 5 2 8o d ' te» ~ ( ) ' C ' Sy ~ ( ) 2a;+a te+ 2a» ' 2x -2x . x +y _ ( ) > . x -a _ ( ) # g ' I - x 2 (1 + x) ( ) ' ' x -y y -x ' x - b 6» - x* 36 Introductory Topics Sec. 1-25 2. Reduce each of the following fractions to lowest terms. 102010 . 8a 2 b 3 6s 4 y 8 350470 * 12a<6 ' C# 20* V ' 36x 4 2/*s 3 48x 5 ?/ 8 f x A - 6x 2 96x 3 y 7 z 2 * e# 24xV 4- 40x 4 ?/ 5 ' 3x 3 - 18x " a 4-26 . x 2 y* - x 3 y 2 . 9x 2 - 49 4a 2 + 8a6 * a; - y U 3x 2 + 13x + 14 ' 2^2 — 7a; - 15 , 6a: 2 - 5x - 6 , x* - 81 k. J ' x 2 - 25 4x 2 + lOx - 24 x 6 + 729 X 2 z (y „ 3)2 fl a _ 86 3 a; 3 - 27y 3 m - 4 — (x — 3/) 2 ' "' a 2 - 4& 2 # °* a: 2 + 3sy + 9y 2 ' a; 2 - (a 4- 6)x + <?& (x + y) (x - y) (y - z) , P * [(2a: 4- a) 2 - (* + 2a) 2 ] (a: 2 - 6) ' q * (?/ 2 - a; 2 ) (z - ?/) 18x 2 4- Sxy - 10y 2 t 9a 2 + 6a6 4- & 2 . f# 21x 2 - 26x2/ 4- 8y 2 ' S * 6a 2 - a& - 6 2 * . x 2 +xy x 2 4- 2xy a: 3 4- a: 2 y 4- a^ 2 4- y 3 ' x 2 + x + 2y - 4a/ 2 1-26. ADDITION AND SUBTRACTION OF FRACTIONS The sum of two fractions with the same denominator is the fraction whose numerator is the sum of the numerators and whose denominator is the given common denominator. That is, in (1-25), (1-25) S + h^r*' For example, x + 2y 3 + 3 3 To add fractions with different denominators, first change the fractions to equivalent fractions having a common denominator, and then write the sum of the new numerators over this common denominator. Ordinarily, the common denominator which is chosen is a least common multiple of the given denominators, since this leaves the fewest possible common factors in the numerator and denominator of the resultant fractions. However, the same result is obtained, after concellation of all common factors, no matter what common denominator is used. The method of finding an L.C.M. was devel- oped in Section 1-23. The difference of two fractions, ~ — ^ > has been defined in o a Section 1-2. Thus, by (1-10), when neither b nor d is zero, < l -»> f-5-f+C-a)- Sec. 1-26 Introductory Topics 37 Applying (1-22), we have C" 52 ' 5-5 = 5 + ^- The following procedure is suggested for adding (or subtracting) fractions. 1. Find a least common multiple of the given denominators. 2. Change each fraction to an equivalent fraction having the L.C.M. as the denominator in the following way: For each fraction, note which factors are in the L.C.M. but not in the denominator of the given fraction. These factors may be found by dividing the L.C.M. by the denominator of the given fraction. Then multiply the numerator and the denominator of the given fraction by these factors. 3. Write the sum (or difference) of the numerators of the new fractions found in step 2 over the L.C.M., and reduce the resulting fraction to lowest terms. The following examples will indicate a procedure which should be followed until some skill in working problems has been attained. 3 5 7 Example 1-35. Express j - -r -f - as a single fraction reduced to lowest terms. Solution: Step 1. The methods in Section 1-23 give us 36 as the L.C.M. of the denominators. Step 2. Divide the L.C.M. successively by the denominators 4, 6, and 9 to get 36/4 = 9, 36/6 = 6, and 36/9 = 4. Change the given fractions to fractions having 36 as the denominator in the following way: 3i?_27 5^6 = 30 7 - 4 28 4 • 9 ~ 36 ' 6-6 36 ' 9-4 36 ' Step 3. Combine the new fractions to obtain 3_5,7_27_3028 27 - 30 + 28 _25 # 4 6 + 9 ~ 36 36 + 36 36 36 ' Note that adding (or subtracting) the numerators and the denominators of the given fractions leads to absurdities. Thus, it is not true that = + o = * * Also, dropping the common denominator leads to absurdities. Thus, it is not true thit I + I-- : 3x 2 5x — 2 Example 1-36. Express — 7-3 - -5 i i as a sm gl e simplified fraction. 2 2 Solution: Step 1. Since x 2 - 4 = (x + 2) (x — 2) and — ^ = 5 > an 38 Introductory Topics Sec. 1-26 L.C.M. of the denominators is x 2 - 4. The given expression may be written as follows: 3x 2 5x - 2 x + 2 x - 2 x 2 - 4 Step 2. When the L.C.M. is divided by the denominators fa + 2), fa — 2), and (x 2 — 4), the results are x 2 - 4 _ x 2 - 4 __ a: 2 - 4 _ , ¥+T - * - * ^T ~ x + ^ o^l - x - Multiplying the numerators and the denominators of the given fractions by these factors, we obtain 3x * fa - 2) _ 3x 2 - 6x (x + 2) • fa - 2) ~ x 2 - 4 ' 2 • fa + 2) 2x +4 (s - 2) • (a + 2) ~ x 2 - 4 ' 5s - 2 _ 5x - 2 _ x 2 - 4 "~ x 2 - 4 ' Step 3. Therefore, the desired result is obtained in the following way: 3x 2 5x - 2 _ (3x 2 - 6x) + (2x 4- 4) - (5x - 2) a? + 2 ' s - 2 a: 2 -4 x 2 - 4 3x 2 - 9x 4- 6 3 fa - 1) (a; - 2) __ 3fa - 1) # ~ x 2 - 4 "" x 2 - 4 x 4- 2 ' EXERCISE 1-8 Perform each of the indicated operations and express the answer in lowest terms. 4 1 + 5 - A 8 + 6 12 ft 5 17 ,19 1. 1 2 7 2 3 + 8* 3. 3-5+1. 7 5+ 21 5. 3+ 20 5 10 ' 7. 1 1 1 + X x — 1 9. 3x - 2 x - x z - 1 'l- -4 - 1 11. is 1 £ _ 6 bi a: x 2 x 3 ' 1 2 + ■ 8. 2 13 1 - x 1 4- z 1 - z 2 10. 2x Sx 2 x 4- 2 x - 1 + *' 12. 1 1 - * - x -1 ' - 1 -X 14. x - 2 x + 2 8x x + 2 2 - x 4 - x 2 !, x 4-3 3-1 X 2 - 1 ^fa - l) 2 x -2 x4-2 x 2 + 5x 4- 6 "*" x 2 4- 7x 4- 12 ^ x 2 + 6x 4- 8 M. » - 2 +1. 17.1+. + - 2 * x 2 — 3x x — 3 x 1 — x 18 g 4- g + 1 - 2x * ~" 3 . 19 2x Sx x + 1 x x 2 + x x 2 - y 2 x 2 - 2xy + y 2 Sec. 1-27 Introductory Topics 39 3 1 20, 4 - x 2 x - 2 22.-JL..+4+: 6 * x - 1 2 - x 24. _* y —+ xy ■ # - 2/ & - y x 2 - y 2 Sx 2x 26. OJ a -f 6 a ~ 6 a — 6 a -f- 6 2ab a 2 - b 2 23.i+? + A. x y xy 25. 1 + ' x + y x - y xy x 2 - y 2 4xy x 2 — y 2 x 2 — 2xy + y 2 x 3 — x 2 y — xy 2 -f y 3 1-27. MULTIPLICATION AND DIVISION OF FRACTIONS The product of two fractions is the fraction whose numerator is the product of the numerators and whose denominator is the product of the denominators of the given fractions. That is, by (1-29), if neither 6 nor d is 0, , oqn a c _ ac (1 ~ 29) Vd'bd 9 To illustrate, 2 6 _ 2j_6 _ 4 , x - 1 x + 3 _ x 2 + 2x - 3 3 ' 7 ~ 3 ■ 7 ~~ 7 ' and z + 2 ' z - 2 ~ x 2 - 4 A special case of (1-29) is worth noting. To multiply a fraction by a number or expression, we multiply the numerator by that number or expression. Thus, b _ a b __ a • b _ ab c ~" I c "" 1 • c ~~ c For example, , 1N 2 • - = - > and (x — 1) • - = — • 5 5 J x x The quotient of two fractions, T / -. > has been defined in bi a Section 1-2 and evaluated in Section 1-7. Thus, if no factor in a denominator is zero, /, Q m a /c a d ad (1 ~ 30) &/d = 6'c = te - To illustrate, 5 /3 __ 5 2 _ 5j_2 _ 10 . 3x 2 /tef __ 3z 2 j£_ _ 3a; 2 * y 7 _ j£_ 7/ 2 ~ 7 ' 3 " 7 • 3 "" 21 ' y 3 I y 7 " y z ' 6x 5 y z • 6z 5 2x*' In practice, we divide out all factors common to the numerator and denominator before proceeding with the actual multiplication. If necessary, the numerator and the denominator of each given fraction should be factored. %2 2x 2x 2 Sx y Example 1-37. Multiply . — —= by - s and simplify the result. 40 Introductory Topics Sec. 1-27 Solution: The work may be indicated as follows: x 2 - 2x t 2x* - 3s - 9 _ x(x - 2) (x - 3) (2s 4- 3) 2x* + 5s + 3 ' s 2 - 9 ~ (s 4- 1) (2s 4- 3) (x + 3) (a; - 3) __ j;(3; - 2) __ x 2 - 2s "" (s 4- 1) (s 4- 3) ~~ x 2 4- 4s 4- 3 ' w i i oo 7^- -j ^ 2 - 5s + 4 , s 3 - 4s 2 -f s - 4 Example 1-38. Divide^— 3^-^ by ^-^ Solution: By (1-30), s 2 - 5s + 4 s 3 - 4s 2 4- s - 4 __ (s 2 - 5s -f 4) • (2s - 1) s 2 - 3s - 4 ' 2s - 1 (s 2 - 3s - 4) • (s 3 - 4s 2 4- s - 4) - (s - 1) (s - 4) (2s - 1) ~ (s + 1) (s -4) (s -4) (s 2 +l) (s - 1) (2s - 1) (s + 1) (x - 4) (s 2 + 1) 1-28. COMPLEX FRACTIONS The fractions we have been discussing so far may be called simple fractions, to distinguish them from the fractions which we now discuss. A fraction which contains other fractions in the numerator or in the denominator is called a complex fraction. Since the simpli- fication of a complex fraction is essentially a problem in division, we first reduce the numerator and the denominator of the complex fraction to simple fractions and then proceed as in division. Example 1-39. Simplify t r • s y First Solution: The numerator of the given complex fraction reduces to the simple x -4- v V 4~ X fraction > and the denominator reduces to • Hence, we have s , xy s -f y x __ (s + y) • xy = y+g x • (?/ 4- s) st/ Alternate Solution: Frequently it may be more convenient to multiply both the numerator and the denominator of the complex fraction by an L.C.M. of the denominators of all simple fractions occurring in the given complex fraction. In this y 1 1 example, the simple fractions are - > - > and - > and an L.C.M. of their denomi- i. • xx y nators is xy. Therefore, by multiplying the numerator and the denominator of the complex fraction by xy } we get y\ xy (■+!)■ xy xy +y 2 = (sj- y)y Sec. 1-28 Introductory Topics 41 EXERCISE 1-9 In each of the problems from 1 to 20, perform the indicated operations and express the answer as a simple fraction in lowest terms. , 3 ]8 2 5 # ' 5 ' 155 ' 3 7 . 3 | 4 iy tofi_ % 7 ' 6t/ ' 9x ' 16?y 3 # 5 14 /28 fi Jx_ / 16x 3 g " 9/45* 17t/ 3 / 68?y 6 ?/ 6a 2 6 3 3a 3 6 - 27a; 3 ?/ / 15^ 15a 3 6 5 ' 48a6 2 ' 72r?/ 3 / 16x ' 15a 4 6 4 / 54aft 2 3af J02_ Jty^ 15a 3 / 14a 3 6 2 ' 4?/ ' 27x 3 ' 15z 3 ' n 9a; 4 ?/ 3 a: 2 - 6x + 5 12 x 3 - 8 a 2 + 2s - 3 x 2 - 25 ' 6a; 3 ?/ 2 ' * 9 - x 2 ' x 2 + 2a; + 4 ' x 2 + 2x - 3 x 2 + x - 6 x 4 + 27a; . x 3 ± 3a; 2 - 2a; - 6 (a; - 7) 2 : a; 2 - 5ar - 14 ' a; 4 - 4x 2 ' x(x 2 - 5a; + 6) 27a; 3 - 8 4x 2 - 25 15. 16. 17. 18. 19. 20. 6a; 2 + 19a; + 10 9x 2 - 12a: + 4 6a; 2 + 5a; - 21 6a; 4 1- 36a; 3 2x 2 - 5x - 12 4a; 2 - 9a; - 28 ' 9 - 4a; 2 ' 9a; 3 + 21x 2 2x 2 ± 5x + 3 # 3a; 2 - 20a; + 12 ^ 6a; 2 + 5a; - 6 # 5a; 2 - 24a; - 5 ' x 2 + 3a: + 2 *' 4a; 2 + 9a; + 2 ' x 2 - (2x - 3*) 2 _^ l~ 4x 2 = (3g - a;) 2 9s; 2 - (a; - 2?/) 2 1 (x - 3z) 2 - 4?/ 2 : L(2?y - x) 2 - 9z 2 ' (3z - 2y) 2 - x 2 J 8x 2 - 2x - 15 9 - 4xH 4x 2 + 12x + 9 j" 8x 2 - 2x - 15 ^ 9 - 4x 2 H L 4?/ 8 * 9 *' 36?/ 7 * 7 J 22. ,7 x " • 23. 4?/ 8 z 9 * 36?/ 7 * 7 J 48x 2 ?/ 3 + 60?/ 3 (x 4 - 625) (x 2 - 9) 3x 4 -f 75x 2 (x + 5) 2 (x - 5) 3 : x 2 + lOx + 25 ' In each of the problems from 21 to 38, simplify the complex fraction. 1 _2 3 2 3 + 4 5 ^6 ^8 16x 2 ?y 3 9x 18x 3 y Axy a T7^ """«_l + i' j/ 2 . 6^6* 3 3a T 3 Itt.|IfL. 31.5ll£LL. 32.1^2. 2_2_4 !_ 4 oX 20 a; 8 a; *» a: 2 - 4 91 2 3 3*4' 7 2 421. 5 16 24. 7 18 4 15 -a 1 2 3 5 ♦s 27. 8x 3 - 2x - -2,3 -y ■ • 5 3 9* 10* 18 19 64 81 6x?/ 5 4x 2 2y X -3 25. -ihr • 26. 28. , * „ • 29. 42 Introductory Topics Sec. 1-28 1 x , 1 - 2x x - 5 + x - 3 - 33. £ + *. 34. 2LZ*. 35. ^±1 34. x • -2 - z — 2 x • -4 - x 1 4 z + 2/ a; -2/ 37. £ -2/ z + v a; 2 + y 2 a: 2 -y 2 1 -2a; # -|- 4 x-4 2a? + 1 a: 1-s 1 + a; x +y x - y a; - 2j/ fi 2t/ 1 L 2y - ad *fi 1 + x * ~ g 27 LzJ g + y Qft s + 1 L 2z/ x + l x - l ' a; 2 +y 2 a: 2 -!/ 2 ' aM-Jj/f 1 a - 1 a: + 1 x 2 -y 2 x 2 +y 2 x 2 - 1 a: - 1 Simplify each of the following expressions: , 1 1 a; +- a; a; a; 1 — 1+- a; +- x +- 39. 2 — j — £. 40 # 2L- 41. \ X + r-rr 1 + r x + 1 a; - 1 1111 " ' * -1 V 35 s 4- v 8 J4k a? 3 V s X 2 v 2 42. — i — = — * * ^% • 43. ^ V - ~ *r • i _ i a; 2 -?/ 2 JL __ i_ lo-i a; 2/ a; 2 y 2 a; y *e*i-)(s^-,-k)*(— +^- «-(*-^)(»- a ^^)*( 2 -ifT)- 1-29. LINEAR EQUATIONS Introduction. In Section 1-13, an equation was defined as a state- ment of equality between two algebraic expressions. In this section we shall discuss equations in one unknown of the simplest type, called linear equations, typified by the form (1-53) , ax-b } where a and b are specified numbers and a^O. Some ideas pertain- ing to equations in general are needed first. Solution or Root of an Equation. We shall use the term unknown to designate a literal quantity that appears in an equation and is not regarded as specified at the outset. A system of values of the unknowns which, when substituted for them, makes the equation a true assertion is called a solution of the equation. A solution is also called a root when only one unknown is involved. Thus, the values x = 2 and y = — 1 define a solution of the equation 3# + 2# = 4, since the equation is satisfied for these values ; that is, 3(2) + 2(-l) = 4. Sec. 1-30 Introductory Topics 43 Equivalent Equations. Two equations are equivalent if they have exactly the same solutions. For example, 2x + 11 = Ix - 4 and Ax + 22 = lAx - 8 are equivalent, since, as will be seen, each has exactly one root, namely, x = 3. However, 2x + 11 = Ix - 4 and 2x 2 + llz = 7z 2 - 4# are not equivalent, because the latter equation has, in addition to the root x = 3, the root x = 0. Operations on Equations. Each of the following operations on an equation yields an equivalent equation : Adding the same expression to (or subtracting the same expres- sion from) both sides. Multiplying (or dividing) both sides by the same non-zero number. For any solution of F = G also makes F +H=G+H true, and is therefore a solution of this latter equation. Conversely, any solution of F + H - G + H is one of F = G, because F = F + H-H = G + H-H = G. Likewise, any solution of F = G is one of aF = aG, provided that a is a non-zero number, and conversely. Transposition of Terms. Transposing a term of an equation con- sists in moving the term from one side of the equation to the other and changing its sign. This operation is equivalent to adding the same quantity to both sides (or subtracting the same quantity from both sides) . For example, consider the equation 2x+ 11 = 7x - 4. If we transpose 2x from the left side to the right, and transpose -4 from the right to the left, we obtain 11 + 4 = Ix - 2a?. In effect, we subtracted 2x from each side and added 4 to each side. 1-30. LINEAR EQUATIONS IN ONE UNKNOWN An equation in the form shown in (1-53) is called a linear equa~ tion in one unknown. We shall show that such a linear equation has one and only one root, namely, (1-54) x = - • a If each side of the equation ax = b is divided by a, the result is b x = - > a 44 Introductory Topics Sec. 1-30 which is equivalent to the given equation. Therefore, (1-53) clearly has one and only one solution, namely, b/a. It should be noted that if we allow a to be in (1-53), there are two possibilities. Either no solution exists, when 6^0, since for no number x is it true that a*x = 0'X = = b¥=0; or else every number # is a solution, when 6 = 0, since a*# = 0*# = = & for all values of x. An equation which is not apparently linear may frequently be solved by the theory of linear equations, by replacing the given equation by a linear equation to which it is equivalent. The follow- ing steps serve as a guide to the method to be used when there is only one unknown in the equation. 1. Clear the equation of any fractions with numerical denomina- tors by multiplying both sides of the equation by a least common multiple of the denominators of those fractions. 2. Transpose all terms containing the unknown to one side and all other terms to the other side. We may collect the terms contain- ing the unknown on either side. 3. Combine like terms. If the equation now assumes the form in (1-53), it is a linear equation and can be solved by dividing both sides by the coefficient of the unknown. 4. Check the result obtained in step 3 by substituting in the original equation. While it is desirable to include step 4 to show that the number x found in step 3 is actually a solution of the equation, step 4 is not a necessary part of the solution process, since the operations performed in the preceding steps always yield equivalent equations. The purpose of step 4 is to help to make certain that there has been no error. In step 2, we generally transpose terms containing the unknown to whichever side makes solving the equation easier. The following explanations should be noted carefully. Students at times "transpose" coefficients of the unknown. Thus, 2x = 6 takes the erroneous form x = 6 - 2 by "transposing" 2 to the right side. The correct procedure is to remove the coefficient 2 by division, since it is a multiplier of x. Therefore, we should divide both sides of the equation 2x = 6 by 2 and have 2x 6 2" = 2' or x = 3. Errors of this type may be avoided if the student applies the rules of algebra properly and checks his solutions carefully. Sec. 1—30 Introductory Topics 45 Multiplying or dividing both sides of an equation by a poly- nomial involving the unknown will not necessarily yield an equiva- lent equation. When the operation is multiplication, the new equa- tion thus obtained may have roots in addition to the roots of the original equation. These extra roots are called extraneous roots. We then say that the equation is redundant with respect to the original equation. When both sides of an equation are divided by a polynomial involving the unknown, the new equation may lack some of the original roots. It is then said to be defective with respect to the original equation. If both sides of an equation are multiplied by a polynomial involving the unknown, the check in step 4 of the recommended procedure is a necessary step in the solution process. An extra- neous root can thus be identified. Dividing both sides of an equa- tion by a polynomial involving the unknown is not a permissible procedure, since roots that are lost cannot be regained. It should be noted that a non-linear equation may some- times be treated so as to obtain a linear equation which is possibly redundant. For example, the fractions in the equations of Problems 39 to 48 of Exercise 1-10 may be eliminated by multiplying both sides of each equation by a least common multiple of the denomina- tors of the fractions in that equation. Since this multiplier con- tains the unknown, the new equation may have solutions which are not solutions of the given equation. Hence, the solutions must be checked to see whether or not they actually are solutions of the given equation. 2^ x + 7 Example 1-40. Solve the equation — = 4 — — Solution: We clear the equation of fractions by multiplying both sides by 15, which is an L.C.M. of the denominators of the fractions, and obtain 10* = 60 - 3x - 21. Collecting the terms containing x on the left side and all other terms on the right, we have 10* + 3* = 60 - 21. Combining like terms, we obtain 13x = 39. Finally, dividing both sides by 13, we have ' x = 3. To check, substitute 3 for x in the original equation. The result is 2l2 = 4 - L+Z , or 2=4-2, or 2=2. Therefore, 3 is the root. 46 Introductory Topics Sec. 1-30 Example 1-41. Solve the equation 6x — 3y — 1 = by + 2x + 11 for y in terms of x. In this case, regard x as specified. Solution: Collect the terms in the unknown y ) on the right side, and collect all other terms on the left. The result is 6x - 2x - 1 - 11 = 5y + 3y. Combining like terms, we have 4z - 12 = %. Now we divide both sides by the coefficient of the unknown to obtain 4x - i2 — — = »• Then ' 4* -12 x-3 V =— — =-2— x — 3 To check, substitute — ~ — for t/ in the original equation, and obtain This equation reduces to 12x - 3(x - 3) - 2 = 5(x - 3) + 4z + 22, which becomes 12z - 3a; + 9 - 2 = 5x - 15 + Ax + 22, or 9x + 7 = 9x + 7. x — 3 Since this result is an identity, y = — — is the solution of the original equation, regardless of the value of x. Example 1-42. Find two consecutive integers such that four times the first is equal to six times the second diminished by 20. Solution: Let x be the smaller integer, and x -f 1 the next larger integer. Then, from the statement of the problem, we have 4x = 6(x 4- 1) - 20. From this equation, we obtain 4x = Qx + 6 - 20. Then, 14 = 2x, or x = 7. Hence z = 7 and x 4- 1 = 8 are the two consecutive integers. The student should carefully check these values by substitution in the original statement of the problem. Example 1-43. The speed of an airplane in still air is 400 miles per hour. If it requires 20 minutes longer to fly from A to B against a wind of 50 miles per hour than it does to fly from B to A with the wind, what is the distance from A to B1 Sec. 1-30 Introductory Topics 47 Solution: Let x be the distance from A to B. From the data, the speed of the plane against the wind, in miles per hour, is 400 — 50 = 350, and the speed of the plane with the wind, in miles per hour, is 400 + 50 = 450. Since distance = rate • time, or d = rt, we have - = t. Hence,, x ;jrrr = time, in hours, required to fly from A to B, and x jrjr = time, in hours, required to fly from B to A. Therefore, from the statement of the problem, it follows that x x __ 1 350 "450 "3* Solving this equation, we have 9x - 7x = 1050, or x = 525. So the distance from A to B is 525 miles. EXERCISE 1-10 Solve for the unknown in each problem from 1 to 15. 1. - 4x - 2 - 3 - 2x. 2. 3x - 6 = x + 12. 3. 3x - 2 = - 4z - 5. 4. Sy + 7 = 2 - 2y. 5. - 9x - 7 = 6 - — « 7. 4 - 3z = 6(1 + 2s). 8. 4x - 6 = 3x + 4. 10. 62/ + 7 = 52/ + 6. 11. 10z - 3 = 9x + 4. 13. 5z - 7 - Sx = 4z - 17 - 6z 14. lty + 4 + 6y = 2y + 7 + 3y. 15 Solve for y in terms of x in each problem from 16 to 24. 16. 2x - y = 3. 17. 23z + 2y = 4. 18. a: - 4?/ + 10 = 0. 19. 4y - 2z + 2 = 0. 20. - + f = 1. 21. 6x - 2y = 3. 0,0 A 22. 3(x - 4) + 4y = - 3. 23. 2a? + 4(y - 3) = 5. 24. 3x + 7(y - i) = | . Solve for all values of x which satisfy the equation in each problem from 25 to 48. 25. 5x - 3 = 4(s - 2). 26. 8te - 2^ - 9(* - 4) = 13. 27. ^±1 = n . °~ ^ 17 9 - 4z _ 4 3Q 5 +6a; _ _ 5 6. 4w + - = 3»-|. 9. a? - 8 = : 2s + 3. 12. \x - 4 5 2 : 3 1 = — x • 2 4 2?. 5 + 4x) - 3(* - 4) = 0. 28. 3 _ 7. „ a; +5 _s +7_ 4 5 29. 32. OJ 2a: - 1 a; — 5 34 - 3 = 7 • 35. „ 8* - 21 , 1 _ 5x 5 4 6 »s-i = e(-J). 6 3 ""' 3 x - 6 _ s - 8 »„ x +3 _ 2a; - 7 "T"-^^' 33 -~5~-~T2~' 2a; +3 3a: - 1 7a; +3 2 , 1 ~4 2~ =3 - 36 --6- = 5 X+ 2' 7 - x 4x + 3 6* 16 38. 4 7 5 3 40 -^2x +3 = 7 - 48 Introductory Topics Sec. 1-30 * + x 1 x x' ' x 9 9 x ' ,„ 4 5 1 ., 1 . 3 4 2x + 3 + x - 4 4x* - lOx - 24 3 1 45 * 6x2 _ 2x + 1 ~ 2^17+7 = °- 46, 47. ^ 7 ~* x z - 1 """# + 1 a? - 1 19 - 22# a: + 1 x +2 z 2 - z - 6 x - 3 5 - 2x - .t 2 6(x - 3) T 2(x 2 + 3x + 9) 81 - 3x 3 48> 4 + 4+4»_ -5 x x 2 + 4:X x + 4 49. Divide 98 into two parts such that one of them exceeds the other by 18. 50. Find three consecutive integers whose sum is 84. 51. Find two consecutive integers whose squares differ by 13. 52. If 8 times a certain number is 9 more than 5 times the same number, what is the number? 53. A rectangular plot of ground is four times as long as it is wide. If its perimeter is 4,800 feet, what is its area? 54. How many pounds of coffee at 90 cents per pound and how many pounds at 98 cents per pound will it take to make 100 pounds of a mixture costing 96 cents per pound? 55. At a college play, admission was 25 cents for a child and 75 cents for an adult. If $210 was taken in from 500 admissions, how many children and how many adults were admitted? 56. A man has $365 in 41 bills of $5 and $10 denominations. How many bills of each denomination does he have? 57. What are the angles of a triangle, if one angle is three times the second angle and six times the third angle? 58. One man, X, can do a certain job in 7 days, and another man, Y, can do the same job in 15 days. How long would it take them to do the job working together? 2 The Function Concept 2-1. RECTANGULAR COORDINATE SYSTEMS IN A PLANE In Section 1-5 we saw how we can associate a real number with every point on a number scale. The real number attached to a given point is called the coordinate of the point. This representation sug- gests the assumption that to any real number there corresponds precisely one point on the scale, and to any point of the scale there corresponds precisely one real number. This one-to-one correspond- ence between the set of real numbers and points on the number scale is known as a one-dimensional coordinate system. We shall now extend the concept of a one-dimensional coordinate system to a system of coordinates in a plane in which two number scales are perpendicular to each other. The two perpendicular lines, which we shall call coordinate axes, divide the plane into four parts, or quadrants, numbered as shown in Fig. 2-1. The horizontal and vertical lines are designated as the x-axis and the y-axis, respectively, and their point of intersection is called the origin and is labeled O. H On each of these axes, we construct a number scale by selecting an arbitrary ^ ►JT unit of length and the origin as the zero point. As in Section 1-5, a coordinate on jjj jy the #-axis will be considered positive if it is to the right of 0, that is, to the right of the y-axis, and will be negative f ig# 2-1. if it is to the left. A coordinate on the y-axis will be considered positive if it is above the #-axis, and nega- tive if it is below. Just as the real-number scale of Fig. 1-1 gave us a system of one-dimensional coordinates by which we could set up a one-to-one correspondence between points on a line and real numbers, so the 49 Q ii ♦ i 50 The Function Concept Sec. 2-1 system of coordinates with respect to two mutually perpendicular axes sets up a one-to-one correspondence between points in a plane and ordered number pairs. We use the designation "ordered" pairs for the following reason. To designate any point, we shall agree to give its directed distance frdm the i/-axis first, and call it the abscissa or x-coordinate, and then the directed distance from the #-axis and call it the ordinate or y -coordinate. The abscissa and ordinate of a point constitute its rectangular coordinates. They are written in parentheses as an ordered number pair, as in the nota- tion (x, y)> the abscissa always being written first. By this scheme we assign to each point of the plane a definite ordered pair (x, y) of real numbers and, conversely, to each ordered pair (x, y) of real numbers there is assigned a definite point of the plane. Thus, the abscissa of the point P in y Fig. 2-2 is 3, and its ordinate is 2, and we say that the coordinates of the point P(3,2) P are (3, 2). Similarly, the coordinates T of Q are (—2, 0), the coordinates of R ni l — ►* are (0, —3), and those of the origin are (0, 0). Marking in the plane the posi- . hR tion of a point designated by its coordi- nates is called plotting the point. The coordinate system we have con- Fig. 2-2. stfucted is a particular case of cartesian coordinates, so called in honor of Rene Descartes (1596-1650), who first introduced a coordinate system in 1637. It is called a rectangular system, since the axes intersect in a right angle. (Actually the axes may intersect at any angle, but it is usually simpler to take them perpendicular to each other. When the two axes are not perpendicular, the coordinate system is called an oblique coordinate system. Oblique systems will not be used in this book.) 2-2. DISTANCE BETWEEN TWO POINTS It was seen in Section 1-10 that |a — b\ equals the distance between two points on the number scale represented by the real numbers a and 6. It follows that \x 2 — Xi\ represents the distance between the points A (x u 0) and B(x 2 , 6) on the #-axis of Fig. 2-3. Let us now consider the two points P\ and P 2 with coordinates (x lf y x ) and (x 2 , yi). The points have the same ^/-coordinate, which means that they lie on the same horizontal line. Hence, the dis- tance between the points Px(x u Vi) and P 2 (x 2 ,yi) is the same as Sec. 2-2 The Function Concept iK *2l*vY\) * i « in i — i i i 1 w V A(x v 0) B(x 2 ,0) ~ Fig. 2-3. Fig. 2-4. the distance between A(x lf 0) and Z?(# 2 ,0). This distance is \x 2 - x±\. Similarly, we can show that \y 2 - Vi\ represents the dis- tance between two points (x 2 , Vi) and (x 2 , y 2 ) on a vertical line. We have thus arrived at the following two important properties. If two points Pi(x u y x ) and P 2 (x 2 , yi) have the same 2/-coordi- nate, then the distance between them, or |PiP 2 |, is given by (2-1) |PiP 2 | = |x 3 — xi|. If two points Qi(x lt y x ) and Q 2 (x u y 2 ) have the same ^-coordi- nate then the distance between them is given by (2-2) IQ1Q2I = I2/2 ~ 1/1 1 . The concept of the distance between any two points in a plane is so important that we shall now develop a formula for it. Let us denote by d the distance between the points Pi(x u Vi) and P 2 {x 2y 2/2). That is, d is the length of the line segment PiP 2 in Fig. 2-4. Let P 3 be the point (x 2 , y x ) as shown. Since the angle at P 3 is a right angle, we have, by the Pytha- gorean theorem, IP1P2I 2 = IP1P3I 2 + IP3P2I 2 . Therefore, |PlP2| 2 = |*2 - Xi\ 2 + |tf2 ~ »l| 2 = (a?2 ~ *i) 2 + (2/2 ~ 2/1) 2 . That is, the distance d between any two points Pi (#1,2/1) and ^2(^2, 2/2) in the plane is given by (2-3) <Z = V>2 - *i) 2 + (2/2 - 2/1) 2 . This formula is known as the distance formula. Example 2-1. Find the distance between the points (— 3» 2) and (5, 2). Solution: It makes no difference which point is labeled Pi. Let us label the first one Pi and the second one P 2 . Since the two points have the same ^-coordinate, (2-1) applies and \PiP 2 1 = |5 - ( - 3)| = |5 + 3| = |8| = 8. 52 The Function Concept Sec. 2-2 Example 2-2. Find the distance between the points (3, 1) and (3, 7). Solution: Again let us label the first point Pi and the second one P 2 . Since the points have the same ^-coordinate, (2-2) applies and IP1P2I = |7 — 1| = 6. Example 2-3. Find the distance between the points (2, - 1) and (5, 3). Solution: Let us designate the points as Pi(2, - 1) and /Mo, 3). By (2-3), the distance IP1P2I is d = y/(b -2)2 +(3 -(-l)) 2 = V9 + 16 = V25 = 5. We shall now consider a special application of the distance formula (2-3). The line segment OP from the origin to a point P(x,y) is called the radius vector to P. By (2-3), the distance between O(0, 0) and any point P(x, y), or the length of the radius vector to P, is or d = V(x - 0) 2 + (y - 0) 2 (2-4) d = V* 2 + 2/ 2 - Thus, for the point P(3, 2), the radius vector has the length EXERCISE 2-1 1. Plot the following points: a. (3,5). b. (-4,7). c. (5, -2). d. ( - 3, - 6). e. (1, - 3). f. ( - 8, - 6) g. (0,2). h. (-5,0). i. (3,0). 2. In each of the following cases, plot the pair of points and find the distance between them: a. (2, 3) and (7, 3). b. (5, - 2) and ( - 1, - 2). c. (1, 4) and (1, 0). d. ( - 3, - 2) and ( - 3, 4). e. (2, 1) and (5, 6). f. (0, - 8) and (5, - 3). g. (3,2) and (5, 7). h. ( - 4, - 6) and ( - 8, - 6). 3. Find the length of the radius vector to each of the following points : a. (4, 3). b. (12, 5). c. (1, - 1). d. (5, -12). e. (7,0). f. (-3, -2). g. (0, 4)._ h. (-1,2). i. (a, 6). j. (W3). k. (-\/3, -1). 1. (ro,n). 2-3. FUNCTIONS So far we have been concerned with single numbers and pairs of numbers. Now we shall consider mathematical relations, known Sec. 2-3 The Function Concept 53 as functions, between two sets of numbers. To distinguish between the two sets, we shall call one of these sets the domain of definition X of the function, and the other the range set Y of the function. We begin by defining a variable to be a symbol which may take any value in a given set of numbers. If # is a symbol which is used to denote any number of the domain X and y is a symbol which denotes any number of the set Y, then x is called an independent variable and y is called a dependent variable. If a set contains only a single number, the symbol used to represent that number is called a constant To set forth a function, the domain should be explicitly specified ; that is, it is necessary to determine definitely just what elements or numbers the domain contains. The same is true of the range set Then, as soon as a definite rule of correspondence is given which assigns to each number x of the domain one or more numbers y of the range set, the function is specified completely. We thus have a set of ordered pairs of numbers (x, y), where x is any number of X and y is a number of Y. The set of ordered pairs of numbers (x 9 y) is called the function. The rule of correspondence which determines the collection of pairs (x,y) is often expressed by a formula involving algebraic or other processes. In such cases we usually find it convenient to refer to the formula as though it were the function. For example, we often speak of the function y = x 2 — Sx + 5 when actually we mean the set of ordered pairs (x,y) determined by y = x 2 — Sx 4- 5. If just one number of Y is paired with each value of x, the function is said to be single-valued. If more than one number of Y is paired with some value of x, the function is said to be multiple-valued. We fre- quently find it possible to deal with multiple-valued functions by separating them into distinct single-valued functions. The range of values of the function consists of those numbers y in the range set Y which actually correspond to some number x of the domain. When the range of a function has been determined, it is always possible to replace the original range set, which may include numbers in addition to those of the range, by the range itself. We shall now consider several examples of functions. Illustration 1. The constant function y = c associates the same number c with every number x of the domain X. Hence, the range Y of the function consists of just one number c. Since y has the same value for all pairs (x, y), the function is evidently single- valued. Illustration 2. The identity function y — x associates with every real number x the number itself. In other words, the numbers 54 The Function Concept Sec. 2-3 corresponding to x = 1, 2, 3, • • • are y = 1, 2, 3, • • respectively. The domain X is the set of all real numbers, and the range Y is also this entire set. The function y = x is single-valued, because it asso- ciates just one number of Y with each value of x. Illustration 3. Let us consider the linear function defined by the equation y = 3x — 2. The domain X is the set of all real numbers, and the range Y is also this entire set. In this case a given x deter- mines a unique y which is equal to Sx - 2. For example, corre- sponding to x = 1, 2, 3, we have y = 1, 4, 7. The function is single- valued. Illustration 4. Let the rule of correspondence be given by the equation y = x 2 . Also, let X be the set of all real numbers, and let Y be the set of all non-negative real numbers. Then, to the number x = -2 there corresponds the number y = (-2) 2 - 4; to the number x = 3 there corresponds the number y = 9 ; and so on. Hence, y = x 2 associates just one number of Y with a given value of x, and defines y as a single-valued function of x. Although this is not obvious, the range of the function is the given range set Y. Illustration 5. In this case let the rule of correspondence be given by the equation y 2 = x. Here X is the set of all non-negative real numbers, and Y is the set of all real numbers. Then to the number x = 2 there correspond the two numbers y = \/2 and y = — \/2 ; to the number x = 9 there correspond the numbers y = V9 = 3 and y = — \/9 = —3 ; and so on. Thus, y 2 = x defines t/asa two-valued function of x. Only for x = is there a single value of y, namely, y = 0. Illustration 6, The function y = y/a 2 — x 2 , with domain X con- sisting of all real numbers x such that — a ^ x ^ a, is a single- valued function with range set Y consisting of the numbers ^ y ^ a. Here the range is Y, as may be proved. Note that y/a 2 - x 2 has no meaning when a 2 — x 2 is negative. Hence, those values of x must be excluded for which a 2 — x 2 < 0, and we must restrict the value of x to the interval — a ^ x ^ a. (We shall consider the meaning of the square root of a negative number in Chapter 11.) The function y = y/a 2 — x 2 is single-valued because y/a 2 — x 2 represents the non-negative square root of a 2 — x 2 . (The other square root of a 2 — x 2 is denoted by —y/a 2 - x 2 .) Illustration 7. The function y = 1/x is defined for all real num- bers different from 0. For x = 0, 1/x is not defined, since division by zero is not permissible. In other words, although there are Sec. 2-4 The Function Contept 55 values of y for values of x neat 1 0, there is no possible value for y when x actually equals 0. Usually the functions which we are about to consider are defined for all values of x, with the following two exceptions : Values of x must be excluded which involve even roots of negative numbers, since these are not defined as real numbers. Values of x must be excluded for which a denominator is zero, since division by zero is not a permissible operation. On occasion, the use to be made of a function will restrict the values of x for which it is to be regarded as defined. 2-4. FUNCTIONAL NOTATION Since functions are mathematical entities, they may be given letter notations, such as /, g, <J>. To designate the number, or num- bers, y corresponding to a given number x according to the rule specified by a given function /, we us6 the notation /(#). As an illustration, let the function / be defined by the equation y =* x 2 — 2x + 3. Then /(0) = 3, /(-l) = 6, and so on. frequently, the symbol f(x) is used to designate the function rather than the functional values. The context will make the meaning cleai*. It should be remembered that the notation y = f(x) does not mean that y is a number / multiplied by another number x. Instead it is an abbreviation for "f of x." The set X does not have to be as simple as in the preceding illus- trations. If X should consist of a set of ordered pairs of numbers, the rule of correspondence would then determine a value, or values, of y for each ordered pair of X. We would then have a function of two independent variables. For example, the area of a triangle is given by the relationship A = /(a, b) = gab. Here X is the set of ordered pairs, (a, b), of positive real numbers, where a is the length of the altitude of the triangle and 6 is the length of its base ; and Y is the set of numbers A, each of which represents an area corresponding to a given pair (a, 6). Similarly, a function of three variables, f(x, y, z), is defined in terms of a set X of ordered triples (x,y,z). Thus, if f(x, y) = x 2 + y 2 , then /(2, 3) = 2 2 + 3 2 = 13. Also, if f(x, y> z) ~ x - y + 2z, then /(3, 2, 5) = 3-2 + 2(5) =11. 56 The Function Concept Sec. 2-4 EXERCISE 2-2 By using the phrase "a function of" in each problem from 1 to 8, express each given quantity, which is regarded as a dependent variable, as a function of one or more independent variables. Where possible write the relationship both in words and in symbols. I. The area of a circle. 2. The area of a triangle. 3. The area of a trapezoid. 4. The volume of a sphere. 5. The volume of a cylinder. 6. The retail price of food in a grocery store. 7. The annual premium for a life insurance policy. 8. A person's height. 9. Given /(as) = 2x - 3, find /(0), /(- 1), /(3), /(1/2), /(V§), 3/(1), /(3)/4, 1 'fix)' 10. Given g(x) = 3x 2 + 5, find g(l), g(- 3), (p(2))«, g(4), 2?(4), <;(* - 1). /( — 2) II. Using f(x) and g(x) as defined in problems 9 and 10, find — y^r- > /(6)gf(3), In problems 12 to 26, let /(#, 2/) = 2xy + 3.r - 2?/, and let g(a> 6, c) = a 2 -f 6 2 + c 2 . Evaluate or simplify each given expression. 12. /(I, 2). 13. /(O, 0). 14. gf(0, 0, 0). 15. </(l, 2, 3). 16. /(*, 1A). 17. 0(p, <?, r). 18. fix, y) + gix, y, ,). 19. A°L . 20 . /^g . /(if,*) . /(</),/© 21. /(I, - 1) • gW% V3, V5). 22. ?(& x, z) 2 %(3,2,1) rf-3,-2,-1) **' 9( a > b > C) - 25. #(- a, b, c) - #(a, - b, c). 26. #(a, 6, - c). 27. Express the area A and circumference of C of a circle as functions of the radius r. By eliminating r, express A as a function of C, and also express C as a func- tion of A. 28. Express the volume of a sphere as a function of its surface area. 29. Express the surface arfea of a sphere as a function of its volume. 30. Suppose that U is a function of V and that V is a function of W. Show that U is a function of W. Determine the maximal domain of values of x for which y is defined as a function of x in each problem from 31 to 61. Assume that x and y are real numbers. 31. y = a. 32. i/ = 3z. 33. 2/ = - | • 34. y = 3z + 1. 35. 1/ = 2a; - 3. 36. 2/ = 4x + 5. 37. 2/ = z 2 . 38. 2/ = z 2 - 2. 39. 2/ = z 2 + 1. 40. y = *(2:r + 1). 41. y = (3s - 1) (a? + 1). 42. ?y = (2x - 3) (2a? + 1). 43. y = x(s - 1) (x + 1). 44. 2/ = z 2 + - • 45. y = g * s(x - 1) 46. x 2 + 2/ 2 = 9. 47. x 2 - y 2 = 9. 48. z 2 - y 2 = 0. z 2 49. x 2 + y 2 = 0. 50. 3s 2 + 2/ 2 = 0. 51. y = 1 + z 2 Sec. 2—6 The Function Concept 57 1 11 x 2 52. y = ^ • 53. y = — — - - - • 54. y = — r-r • * 1 - s * x + 2 x * x 4- 1 55 ' 2/ = *! 7 I * 56. x = 4i/ 2 . 57. x + 1 = 32/ 2 . 58. !/ = Vl - z 2 . 59. y = \/4 - a; 2 . 60. y = \/z 2 + 1. 2-5. SOME SPECIAL FUNCTIONS The following additional illustrations of two rather unusual, but very useful, functions are given here to help us become better acquainted with the function concept. The absolute-value function is defined by associating with x its absolute value \x\. The functional relation is given by y—\x\. Thus, the domain comprises the set of all real numbers, while the range comprises the set of all non-negative real numbers. For this function we have the values 2, 3, 0, ir, y/2 corresponding, respec- tively, to x = 2, —3, 0, 77, — \/2. The bracket function or greatest-integer function, represented by the notation y = [x] , is defined as the largest integer which does not exceed x. Its domain is the set of all real numbers, and its range is the set of all integers. Thus, if f(x) = [x], then /(— 3.5) - -4, /(-l) = -1, /(0) = 0, /(2.5) = 2, and /(5) = 5. EXERCISE 2-3 1. Given /(x) = [x | , find/(- 3),/(2.3), and /(/(*))• 2. Given /(x) = [x], find/(3.2),/(2), and/(- 3). 3. Given /(x) = x r M> find/(- 3) and/(2.5). 4. Given /(x) = | x | + [x] - x, find/( - 3.5) and/(4). 5. Given /(x) = [x] + [2 - x] - 1, find/(0),/(- l/2),/(l),/(3/2), and/(2). 6. Let/(x) be the function whose domain X is the set of all real numbers for which the definition is as follows : if x < 0, then/(x) = - x; if x ^ 0, then fix) = x. Find/(3)and/(-2). 2-6. VARIATION A particularly important example of a simple type of function often occurring in the physical sciences is given by the formula y = kx. If k > 0, this equation shows that y increases a§ x increases, and that y decreases as x decreases. We usually say that y varies directly 58 The Function Concept Sec. 2—6 as x, or that y is directly proportional to x. If x ¥= 0, the relation- ship may also be written as follows : x . ' where k is called the constant of proportionality. This relationship is equivalent to saying that the ratio of y to x remains constant for all non-zero values of x. The value of the constant k in any particular problem may be determined from a known pair of values of x and y in the problem. Thus, if the given relationship is y = kx, and if we know that y = 6 when x = 2, then k = 3. The formula then becomes y = Sx. We say that y varies inversely as x, or y is inversely proportional to a, if h y=l (x * 0). This relationship shows that y decreases as x increases, and that y increases as x decreases. But, when x ¥= 0, the following two for- mulas are equivalent : ^ xy = k and y = - • Therefore, the relationship between a? and 2/ is such that the product of # and y is constant. Several types of variations may be combined in a single equation to express a certain law. For example, when y varies directly as x and z, we say that y varies jointly with x and z, and we write y = te. Direct variation and inverse variation are often combined in applications. Thus, according to Newton's law of gravitation, the force F of attraction between two bodies of masses mi and m 2 varies directly as the product of their masses and inversely as the square of the distance d between th$m. The equation is „ _ kmim2 F--35-. Example 2-4. If a man is paid $15 for an 8-hour day, how much would he make in a 35-hour week? Solution: The wages a man earns vary directly as the amount of time he works. Since this is a problem in direct variation, we have w = kt. In this formula, w represents the total wages, in dollars; t represents the time worked, in hours; and the constant k represents the wage rate in dollars per hour. Substituting w = 15 and t = 8 determine^ the constant k = 1.875. The general formula then becomes w = 1.875 t. Sec. 2-6 The Function Concept 59 Therefore, when t = 35, we have w = (1.875) (35) = 65.62. Hence, the man earns $65.62 in a week. Example 2-5. A motorist traveling at an average rate of 50 miles per hour made a trip in 5 hours. How long would it take him to make the same trip at an average rate of 60 miles per hour? Solution: Since the time required Varies inversely as the speed, we have r In this case, k represents the distance traveled, in miles; r is the speed, in miles per hour; and t is the time required, in hours. Substituting t = 5 and r = 50 determines k = 250. Hence, the general formula is t = — • r Therefore, when r = 60, we obtain 250 _ 25 60 6 * Hence, the time required is 4 hours 10 minutes. Example 2-6. Under suitable conditions the electric current / in a conductor varies directly as the electromotive force E and inversely as the resistance R of the conductor. When E = 110 volts and R = 10 ohms, / = 11 amperes. Find what voltage is necessary to cause 2 amperes to flow through 60 ohms of resistance. Solution: From the statement of the problem, we see that the combined variation is given by the formula ^ 1 "If Substituting I = 11, E — 110, and R = 10 determines the constant k to be 1. Therefore, the formula becomes R This relationship may also be expressed as follows: The required voltage E is equal to the current / flowing through the conductor multiplied by the resistance B, or E = IR. For the specified values, E = (2) (60) = 120. Hence, E = 120 volts. In this problem k = 1. The formula obtained is commonly known as Ohm's law. It is widely applied to entire and partial circuits through which electric currents flow. EXERCISE 2-4 1. If y varies directly with #, and y = 15 wh&i x = 7, find a formula for y in terms of x, 2. If x varies directly with y, and x *= 32 when y = 4, find x When y = 3. 3. If y is directly proportional to x 2 > and y = 112 when a; = 4, find y when a; = 9. 4. If y is inversely proportional to x, and 2/ = - % when x = 1, find y when £ = -3. 60 The Function Concept Sec. 2-6 5. If y varies inversely with x ) and y = 10 when x = 3, find y when a; = 6. 6. If x varies directly with y and inversely with 2, and a? = 4 when y = 12 and 2=2, find # when ?/ = 16 and z = 4. 7. If y varies directly with y/x and inversely with z 2 , and 2/ = 18 when x = 9 and 2=2, find 2/ when £ = 25 and 2 = 6. 8. If t/ is directly proportional to x and inversely proportional to y/z } and 2/ = 4 when # = 1 and 2 = 1, find ?/ when x = 2 and 2=4. 9. If ?/ varies directly with x z and inversely with 1 — 2 2 , and y = 2 when a; = 1 and 2=2, find 2/ when x = — 1 and 2 = — 2. 10. If two spheres have radii n and r 2 , diameters di and ^2, and surface areas Si and £ 2 , respectively, show that r-2 2 ~ d 2 2 ~~ S 2 * 11. If the two spheres in problem 10 have volumes V\ and F2, respectively, show that Vx ^rj* di» 7 2 r 2 3 d 2 3 ' 12. If the radii of two spheres are 3 units and 1 unit, respectively, find a) the ratio of their surface areas and b) the ratio of their volumes. 13. What are the ratio of the surface areas and the ratio of the volumes of two spheres if the ratio of their radii is 3/2? 14. The diameter of the planet Jupiter is approximately 10.9 times the diameter of Earth. Assuming that both planets are spheres, find a) the ratio of their surface areas and b) the ratio of their volumes. 15. The diameter of the sun is approximately 109 times the diameter of Earth. Compare the volumes and the surface areas of the sun and Jupiter, if we assume that the sun and both planets arc spheres. 16. By how much must the diameter of a sphere be multiplied to give a sphere whose surface area is 25 times that of a given sphere? 17. When the volume V of a gas remains constant, the pressure P varies directly as its absolute temperature T. (Absolute temperature is measured from the so-called absolute zero, which is approximately — 460° F or — 273° C.) If gas is enclosed in a tank having a volume of 1,000 cubic feet and the pressure is_ 54 pounds per square inch at a temperature of 27° C, what will be the tempera- ture when the pressure is raised to 108 pounds per square inch? 18. If the temperature T of a gas remains constant, the pressure P varies inversely as the volume V. A gas at a pressure of 50 pounds per square inch has a volume of 1,000 cubic feet. If the pressure is increased to 150 pounds per square inch while the temperature remains constant, what is the volume? 19* The weight of a body above the earth's surface varies inversely as the Square of the distance from the center of the earth. If a certain body weighs 100 pounds when it is 4,000 miles from the center of the earth, hdw much will it weigh when it is 4,010 miles from the center arid when 4,100 miles from the • center? Sec. 2-7 The Function Concept 61 20. The electrical resistance of a wire varies directly as its length and inversely as the square of its diameter. A copper wire 10 inches long and 0.04 inches in diameter has a resistance of 0.0656 ohms, approximately. What is the resistance of a copper wire 1 inch long and 0.01 inches in diameter? 21. What is the diameter of a copper wire 1,000 inches long whose resistance is 10 ohms? 22. According to Kepler's third law, the square of the time it takes a planet to make one circuit about the sun varies as the cube of its mean distance from the sun. The mean distance of the earth is 92.9 million miles, and the mean distance of Jupiter is 475.5 million miles. Find the time it takes Jupiter to make one circuit about the sun. 2-7. CLASSIFICATION OF FUNCTIONS It is often desirable to group functions into classes. For our immediate purpose it will suffice to consider a classification into algebraic functions and non-algebraic, or transcendental, functions. Let us first give a more precise definition of a polynomial function and then define algebraic functions and give some illustrations of both. A polynomial function of x is a function given by the relationship y = a x n + aix n ~ l + h a n _ix + a n , where a , a l9 • • • , a n - l9 a n are real constants, a ¥= 0, and wis a posi- tive integer or zero. The polynomial function is said to be of degree n. The function which makes the number correspond to every number x is also called the zero polynomial, but this poly- nomial has no degree. A rational function of x is a function which either is a poly- nomial function or can be expressed as a quotient of two poly- nomials. Thus, a polynomial is often referred to as a rational integral function of x. A polynomial, or a rational integral function, of x, y, z, ■ • • , is defined to be the algebraic sum of terms of the form kx a y b z c ■ • • , where & is a constant coefficient and each of the exponents a, b, c, •is either a positive integer or zero*. The degree of such a func- tion is the degree of the highest-degree term which is present. For example, the expressions 3x 2 - 5 and 5x 2 — 7xy 2 + &z define polynomial functions of the second degree and third degree, respec- <■»• — in . — oj tively. These and the expressions —~ and 2x — v 7 + 2 T i are examples of rational functions. * Zero exponents will be defined in Chapter 4. For now, one needs only note that u° = 1 for any non-zero number u. 62 The Function Concept Sec. 2-7 A number is an algebraic number if it is a root of a polynomial equation of the form a x n + aix n ~ l + • • • + a n _i# + a„ = 0, in which the coefficients Oo, &i, ■ • • a n are integers, not all zero. Analogous to the term algebraic number, we have the term algebraic function. A function y = f(x) is called an algebraic func- tion of x if y is a solution of an equation of the form Po(x)y n + ^lWr 1 + • • • + P n -i(x)y + P n (x) = 0, where the coefficients P (#), Pi(#), ■ * • , P*(#) are polynomials in x, and n is a positive integer. Polynomials and rational functions are special types of algebraic functions. The functions that we have considered so far were illus- trations of algebraic functions. According to our definition, y = V# is an algebraic function of x because y 2 — x — 0. In this case, n = 2, P (x)=l, P 1 (x)=0, and P 2 (%) = x. Similarly, is an algebraic function of x because xy 2 - x 2 + 1 = 0. y=^ x - X Here n = 2, P (:r) = a?, P^a) = 0, and P 2 (z) = - x 2 + 1. An irrational function is an algebraic function which is not a rational function. A transcendental number is a number which is not algebraic, and a transcendental function is a function which is not algebraic. Functions like the trigonometric functions, which we shall take up in Section 3-2, belong to the class of transcendental functions. Later we shall consider other types of transcendental functions, namely, the logarithmic and exponential functions such as log x and 10*. O The Trigonometric Functions 3-1. THE POINT FUNCTION P(») The trigonometric functions that we are about to define are func- tions in the sense previously described in Section 2-3 ; that is, they are relations between two sets of numbers. The student who is familiar with the trigonometric functions from his high-school work is cautioned to note that we are not, for the present, discussing angles in connection with these functions. We shall see that the concept of a trigonometric function need not be associated with an angle; in fact, many of the most important applications of mathematics in modern science and engineering are concerned with trigonomet- ric functions of pure numbers. Hence, we shall adopt the numer- ical point of view, leaving the study in terms of angles as a secondary consideration. Consider a circle with a radius of one unit placed at the origin of a rectangular-coordinate system. See Fig. 3-1. Let t be any real *»x Fig. 3-1. Fig. 3-2. number. Starting at the point with coordinates (1,0), we lay off on the unit circle an arc of length \t\. If t> 0, we measure the arc in a counterclockwise direction. If t < 0, we measure in the clock- wise direction. If t = 0, the arc consists only of the point (1,0). By this procedure, there is associated with each real number t a defi- nite end-point P(t) of the arc whose initial point is (1, 0). There- 63 64 The Trigonometric Functions Sec. 3—1 fore, corresponding to every real number t, we have a definite ordered pair (x, y) of numbers which are the coordinates of the endpoint of the arc. Since P(t) lies on the unit circle, it is at one unit distance from the origin. Hence, it follows from the distance formula that (3-1) x 2 + y 2 = 1. By means of this equation we can find the second coordinate of the point P(t), except for sign, if one of the coordinates is known. To determine the number pair (x, y) for the point P(t) corre- sponding to a given value of t, we shall take note of the fact that the circumference of the unit circle is 2tt — 6.2832 units (approxi- mately) . For example, since arc ABC in Fig. 3-2 is one-half of the complete circumference, it is tt units in length, and arc AB is equal to it/2. It now becomes apparent that P(0) is the initial point (1, 0) ; P(tt) is the point (-1, 0) ; P(tt/2) is the point (0, 1) ; and jP(3tt/2) and P(-tt/2) both represent the point (0,-1). 3-2. DEFINITIONS OF THE TRIGONOMETRIC FUNCTIONS The correspondence between the set of real numbers t and the set of ordered pairs (x, y) leads to definitions of the six common trigonometric functions, namely, the sine, cosine, tangent, cotan- gent, secant and cosecant. General Relationships. We shall define the cosine of the real number t to be x, and the sine of the real number t to be y. Thus, we have (3-2) x = cos t, and (3-3) y = sin t. The domain of each of these functions (the sine function and the cosine function) is the set of all real numbers t. Since, however, every point P(t) lies on the unit circle, neither of its coordinates (x, y) can exceed 1 in absolute value. Therefore, (3-4) |cos*|^l and |sin*|^l. In other words, the ranges of cos t and sin t are restricted by the requirements —1 ^ cos t ^ 1 and —1 ^ sin t ^ 1, respectively, for all values of t It may be shown that the range of each of these func- tions is the set of all real numbers u such that -1 ^ u ^ 1. The other four functions may be defined in terms of the cosine and sine, as follows : (3-5) tan t = ^4 (cos * * 0), 7 cos t Sec. 3-2 The Trigonometric Functions 65 (3-6) cot t = 5°L? ( sin * * <>)> ' sin £ (3-7) sec * = -^— (cos < ^ 0), COS v (3-8) esc t = -X- (sin * ^ 0). 7 sin £ Since the cosine and sine are defined in terms of the coordinates of the point P(t), it is also possible to express the other functions in terms of these same coordinates. From the definitions of the cosine and the sine given by (3-2) and (3-3) and from the defini- tions of the other functions given by (3-5), (3-6), (3-7), and (3-8), we have (3-9) tan t = ^4 = ~ ' ( x * °)> v ' cos t X (3-10) C ott = ~~ = ~ J (y*0), v ' sm t y (3-11) sec t = -^-7 = - ^ (a? ^ 0), v y cos £ X ~ (3-12) esc * = -J— =- ' (y**0). smty We note here that cos t, or x, appears in the denominators of both tan t and sec t. Hence, tan t and sec t are not defined when t is a number for which the x-coordinate of P(t) equals zero. For example, since the ^-coordinate of P(tt/2) or P(37r/2) is 0, it fol- lows that tan 7r/2, sec 7r/2, tan 3tt/2, and sec 3t7/2 are not defined. Similarly, it can be shown that cot 0, esc 0, cot it, and esc u are not defined. We conclude, therefore, that the domain of each of the functions tan t, cot t, sec t, and esc t is the set of all real numbers for which the denominator is not zero. It also follows from (3-7) and (3-8) that numerical values of sec t or of esc t can never be less than 1. Hence, the ranges of these two functions are restricted by the requirements (3-13) |sec t\ ^ 1 and |csc t\ ^ 1. From (3-5) and (3-6) we obtain an idea of the behavior of the tangent and cotangent functions. It may be shown that the range of each of these funptions is the set of all real numbers. The Trigonometric Functions of 7r/6, 7t/4, and 7r/3. The computa- tion of the numerical values of the trigonometric functions in gen- eral is beyond the scope of this book. However, we shall use the methods of plane geometry to find sin t, cos t, and tan t for £.= 7r/6, 66 The Trigonometric Functions Sec. 3-2 *»x Fig. 3-3. Fig. 3-4. Fig. 3-5. 7r/4, and tt/3, in order to show that for certain values of t the trigonometric functions can be found exactly without tables. To compute the functions for the real value t = 7r/3, we con- struct the unit circle of Fig. 3-3. Arc AB is given to be equal to 7r/3, which is one-sixth of the complete circumference. Triangle OAB is inscribed in the circle, as shown, with side BO extended through the origin to C. Our problem now is to find the values of x = cos (7r/3) and y = sin (tt/3) as coordinates of the point B. Since OA = OB, triangle AOB is isosceles. Hence, angle OAB = angle OBA. We note also that arc CAB = tt and arc CA = arc CAB - arc AB = tt - tt/3 = 2tt/3. Therefore, Also, arc CA = 2 arc AB. angle CO A = 2 angle AOB. Furthermore, angle COA is an exterior angle of triangle AOB. Hence, it equals the sum of the two remote interior angles OAB and OBA, or angle COA = angle OAB + angle OBA. Thus, 2 angle AOB = angle OAB + angle OBA, and triangle AOB is equilateral. If we draw BD perpendicular to OA, it will bisect OA. Then x = 1/2 ; and from # 2 + # 2 = 1 it follows that y 2 = 3/4 and y = VS/2. Hence, cos (tt/3) = 1/2, sin (tt/3) = VS/2, and tan (tt/3) = V& For t = 7r/6, place the equilateral triangle AOB of Fig. 3-3 in the unit circle as shown in Fig. 3-4, where E is the mid-point of arc AB. Then arc EB = 7r/6 and OE is the perpendicular bisector of chord AB. Sec. 3-2 The Trigonometric Functions 67 Since DB is one-half of chordAB, y = 1/2. From x 2 + 2/ 2 = 1 it follows that £ 2 = 3/4 and a = V3/2. Hence, cos (tt/6) = \/5/2 and sin (tt/6) = 1/2. We then have tan (tt/6) = 1/y/S = Vff/3. For £ = 7r/4, construct a unit circle as shown in Fig. 3-5 with arc AB = tt/4. Since arc AC = tt/2, arc 5C = tt/2 — 7r/4 = ir/4. Hence, arc AB = arc Z?C, and angle A 05 = angle 50C. Draw SD perpendicular to OA. Since the two parallels OC and DB are cut by the transversal OB, angle BOC = angle OBZ)., Hence angle AOB = angle OBZ>, and 02) = DB. Thus, y = x. Substituting this value of y in x 2 + y 2 = 1, we have 2# 2 = 1 and # 2 = 1/2. Therefore, a; = cos (tt/4) = \/2/2, and y = sin (tt/4) = V2/2. It follows-that tan (tt/4) = 1. Other Special Values. In a similar fashion we can compute exactly the trigonometric functions of such values of t as — > — > 7tt 3tt — i and — - • Functions of multiples of rr/2 may also be computed 6 4 in this fashion, if one considers a straight line as a right triangle in which one angle is and, hence, one side has zero length. Example 3-1. Calculate the values of the six trigonometric functions of t = ir/2. Solution: As explained in Section 3-1, the coordinates of the point P(ir/2) are (0, 1). Hence, by (3-2), (3-3), (3-9), (3-10), (3-11), and (3-12), cos (t/2) = 0, sec (ir/2) is undefined, sin (tt/2) = 1, esc (tt/2) = 1, tan (tt/2) is undefined, cot (w/2) = 0. EXERCISE 3-1 1. Determine the coordinates of each of the following points: a. P(2tt). b. P(-tt). c. P(5x/2,. d. p(~y)' e - p ( 4 *)- f - p (- 7 *)- 2. In each of the following cases, carefully draw a unit circle and estimate the coordinates of P(t). a. P(l). b. P(2). c. P(3). d. P(-2). e. P(4). f. P(-3). 68 The Trigonometric Functions See. 3-2 3. Evaluate each of the following: . 3tt a, sin -r- • 4 - 57T b. cos -r- • o c. tan(--~) d. COt -r- • . llTT e. sm -=— • 6 f. sec -r- • g. csc (- g)« h. sm (- y)- / 5tt\ .. cob(- t ) 4. In each of the following cases assume that ^ t ^ 2ir, and draw a figure showing approximately the appropriate arc (or arcs). a. sin t = 1/2. b. cot i = — 1. c. tan t = 1. d. csc £ = — 1. e. cos 2 = — 1/2 > sin t being positive. f. cot t = — 1, sec J being negative. 5. Complete the following table, which shows the algebraic signs of the trig- onometric functions in the four quadrants. Quadrant in which P(t) lies Function I II Ill IV cos t + — — ■v sin t + + - -~ tan t + - + — cot t sec t csc t 6. Use the equation x 2 + y 2 = 1 to find all values of t for which tan t = cot t, where £ t g, 2tt. 7. Use the equation x 2 + y 2 — 1 in each of the following cases to find the other trigonometric functions of t. a. sin t = 1/2. b. cos t = 3/5. c. sec t = 13/12. d. csc t = - 3/2. e. sec t = - 2. f. tan i = 4/3. g. cot * = 2. h. tan * = - 6/7. i. sin t = - 3/5. 8. Prove that each of the following equations is correct. a. sin ( - t) = - sin t. b. cos ( - t) = cos t. c. sec ( — t) = sec £. '- d. tan ( — t) = — tan t. 9. For each of the following cases, state the quadrant, or quadrants, in which the given condition is satisfied. a. The sine and cosine have the same signs. b. The tangent and cosine have opposite signs. 3-3. IDENTITIES As an immediate consequence of the definitions of the six trigo- nometric functions, we can establish certain relationships among Sec. 3—3 The Trigonometric Functions 69 them which hold for every value of t. Since P(t) lies on the unit circle, (3-1) holds ; that is, x 2 + y 2 = 1. But, according to (3-2) and (3-3) , x = cos t and y = sin t. Therefore, we have the equation (3-14) cos 2 t + sin 2 t = 1. This states that "the square of the cosine of t plus the square of the sine of t equals unity." Since (3-14) holds for every value of t, it is an identity. Note that we use the symbol cos 2 t instead of (cos t) 2 . This simplified notation is used for all positive exponents, but is never used in the case of a negative exponent. Thus, cos -1 t does not mean the same as (cos t)~\ as we shall see in Chapter 8. Similarly, we can prove that for each value of t for which the functions are defined, (3-15) 1 + tan 2 t = sec 2 t 9 (3-16) 1 + cot 2 t = esc 2 t. Proof of (3-15): By definition, tan t = - — - and sec £ = ■ cos t cos t However, these relationships have no meaning if cos t equals zero, that is, if the ^-coordinate vf P(t) equals zero. When cos t^O, we may divide both sides of the identity cos 2 1 + sin 2 1 = 1 by cos 2 t and obtain sin 2 t ___ 1 „, „ cos 2 t "~ cos 2 t Therefore, 1 + tan 2 t = sec 2 t. cos t 1 Proof of (8-16): By definition, cot t = - — - and esc t = sin t sin J In this case we assume that sin t ^ 0. When we divide both sides of sin 2 1 + cos- £ = 1 by sin- £, we obtain cos 2 2 1 sin 2 t sin 2 £ Therefore, 1 + cot 2 t = esc 2 t. We thus have established the three identities which we restate here for easier reference : (3-14) cos 2 t + sin 2 t = 1, (3-15) 1 + tan 2 t = sec 2 t y (3-16) 1 + cot 2 t = esc 2 L These fundamental! identities are very important in working with trigonometric identities and should be remembered. Our present work with identities will consist of reducing given trigonometric expressions to other forms. Unfortunately, no specific rule of procedure can be given for making these reductions. Profit- 70 The Trigonometric Functions Sec. 3-3 ciency in making such reductions is a matter of both practice and experience. Generally speaking, when we want to reduce a given expression to some other form, it is helpful first to perform any indicated algebraic operations and then to use some form of one of the fundamental relationships to simplify the expression. To prove an identity, we may proceed in any one of the follow- ing ways : 1) We may work on the more complicated member of the identity and attempt to reduce it to the simpler member. 2) We may work with both sides of the identity and show that they induce to the same expression. 3) We may form the difference of the two sides of the identity and prove that difference to be equal to zero. It is frequently desirable to express both sides of the given identity in terms of sines and cosines alone, and then use (3-14) if needed. We shall consider a few examples to illustrate the procedure in reducing expressions. Example 3-2. Show that cos t + sin t tan t = sec t. Solution: From (3-5), or the definition of the tangent, we have * , • * * * 4 . • * sin t cos t -f sin t tan t = cos t -f sin t • Adding, we have . . 9 . , . ., . & * M , . M sin t cos 2 t + sin 2 1 cos t + sin t = • cos t cos t Since cos 2 1 -f sin 2 1 = 1 and r = sec t, cos t cos t + sin t tan t = = sec t. cos t Example 3-3. Reduce — — to sin t cos t. r tan t -f cot t sin t cos t Solution: By definition, tan t = and cot t = - — - • Therefore, J cos t sin t 1 1 1 cos t sin t tan t + cot t sin t cos t sin 2 1 + cos 2 1 sin 2 t + cos 2 £ cos 2 sin £ cos £ sin t Since sin 2 J + cos 2 1 = 1, we obtain 1 cos / sin t tan J + cot t = sin £ cos £. Example 3-4. Establish the identity (sec J - cos t) 2 = tan 2 J(l - cos 2 by reducing both sides to the same expression. Solution: By definition, sec 2 = ; • Hence, the left side becomes cos t Vcos J / \ cos J / Sec. 3-4 The Trigonometric Functions 71 By (3-14) we have x1 9 . 9 , . 9 _ x 9 . . . '1 — cos 2 J\ 2 /sin 2 J\ 2 sin 4 £ / l - cos 2 t y__ / sin 2 l \ 2 _ i \ cos £ / ~~ Vcos t/ ~~ i cos 2 1 The right side of the given identity may be reduced as follows. By definition, . , sin t tan t = - • cos t Hence, we have . n A . M M - 9 ,, t 9 A sm 2 1 . sin 4 J tan 2 J(l - cos 2 t) = — — • sm 2 1 = — — • <?os 2 1 cos 2 1 Since both sides reduce to the same expression, — — > the identity is established. EXERCISE 3-2 Prove each of the following identities: - . , . , , n sin t cos t t 1. sm t - cos t tan t = 0. 2. 7 H = 1. esc 2 sec t sinf secf n 4 . . . . 3. : = 0. 4. tan J esc t = sec £. cos t esc £ 5. sin t(cot t 4- esc = 1 -f cos £. 6. (sin £ — cos 2 = 1 — 2 sin £ cos J. 7. sec 2 J 4- 2 tan t = (1 -f tan 2 . 8. sin £(csc t - sin = cos 2 1. 9. tan 2 *(cot 2 1 - cos 2 = cos 2 U- 10. cos 2 * 4- cos 2 1 tan 2 $ = 1. 11. (1 - sin (1 + sin t) = cos 2 1. 12. (sec t - 1) (sec t + 1) = tan 2 *. 13. — = esc L 14. sin t tan t + cos 2 £ sec t = sec $. 1 - cos 2 £ 15. sin 2 1 esc 2 * = 2 - cos 2 J sec 2 1. 16. sin 4 * sec 2 1 esc 2 J = tan 2 L 17. tan J + cot t = sec J esc J. 18. esc 2 1 4- sec 2 J = sec 2 1 esc 2 J. 19. sec 4 1 - sec 2 * = tan 4 1 + tan 2 1. 20. sin 4 J - cos 4 1 = sin 2 J - cos 2 1. 21. sm * tan * = _ S ec *. 22. tan 2 t + cot 2 « = sec 2 1 csc ? « - 2. cos 2 J - 1 9q 1 — sin ^ 1 — cos < __ sin t + cos I — 1 ^ cos £ sin t ~~ sin £ cos £ , sec 2 f - tan 2 t 4- tan J , . , , 24. = sin t 4- cos L sec £ 25. sec 2 1 = esc 2 *(sec 2 1 - 1). 26. (tan t + cot J) 2 = sec 2 1 esc 2 J. 27. 1 4 tan 2 £ = (sec 2 1 - l)csc 2 «. 28. sin t(\ 4- tan 2 1) - sin * = tan 3 1 cos *. sin t — co s I __ tan t — 1 ^ ~ ft sin 3 6 -f cos 3 t sin < 4- cos t "~ tan £4-1 sin £ + cos £ 1 — cos t 1 + cos t ^ A sm £ — cos t tan f - 1 ^ A sin 3 1 + cos 3 1 - . . . 29. -; — . . = . 30. —: — — = 1 — sin t cos I sm t 4- c" 31. cot 2 1 - cos 2 1 = cos 2 * cot 2 1. 32. * 7 C ° 8 ! = ( csc * - cot 2 • 3-4. TABLES OF TRIGONOMETRIC FUNCTIONS Exact numerical values of trigonometric functions in general cannot be found. However, by use of methods beyond the scope of this book, the values can be computed to as many decimal places as desired. The results of such computations have been tabulated m$ are included in this text in the form of tables of trigonometric functions. Table I at the end of the text contains the values of the 72 I k r Pr V"~ -^*(h) *\f y *i *i V 1 X Jx The Trigonometric Functions Y Sec. 3-4 Fig. 3-6. six functions corresponding to numbers t such that ^ t ^ tt/2. Actually, since it/2 = 1.5708 approximately, the table contains values of £ between and 1.60. Let P(t) = (#, ?/) be the point corresponding to a given value of t, Fig. 3-6, and let t x denote the length of the shorter arc which joins P(t) to the #-axis. In each case in Fig. 3-6, the point P(ti) is located by measuring the arc t x counterclockwise from the posi- tive half of the #-axis. We shall call t x the reference number, or related number, for t. Note that ^ U = tt/2. Since t x is a real number, there is associated with it a point P(t x ) = (x u yi). Also, since t x lies between and tt/2, P(£i) must lie in the first quadrant. In each case in Fig. 3-6 the coordinates of P(U) must be numer- ically equal to those of P{t) ; that is, \x\ — x x and \y\ = yi. Since all the trigonometric functions are defined in terms of x and y, we can say that (3-17) | any function of t \ = same function of t\. These functional values may not have the same algebraic sign, since P(U) lies in the first quadrant and all functions of U have positive values, whereas P(t) may lie in any quadrant and the functions of t do not necessarily have positive values. It is impor- tant to see that the proper sign is chosen to make the equation a true one. The algebraic sign in each case depends on the quadrant in which P(t) lies. The following examples and Fig. 3-7 will illustrate the method of reducing a function of any positive or negative t to the same f unctipn of the reference number t x . We shall limit our discussion for the present to direct use of Table I and consider ipnly values of t x which are shown there. The process of using the table for values of t x which are not shown will be treated in Section 3-10 when we discuss interpolation. For simplicity at this time, we shall use the approximate value 7r = 3.14. Sec. 3-4 The Trigonometric Functions 73 t x -W-2 Example 3-5. Reduce the functions of t = 2 to functions of its reference number. Solution: As shown in Fig. 3-7(a), P(2) is in the second quadrant, and the reference number ti is it — 2, or 1.14. The numerical values of the functions of 1.14 may be found from Table I. The signs of the functions of 2 are determined by noting that only the sine and cosecant are positive in the second quadrant. Thus, we have : cos 2 = - cos (x - 2) = - cos 1.14 = - 0.4176, sin 2 = sin (t - 2) = sin 1.14 = 0.9086, tan 2 = - tan (tt - 2) = - tan 1.14 = - 2.176, cot 2 = - cot (t - 2) = - cot 1.14 = - 0.4596, sec 2 = - sec (x - 2) = - sec 1.14 = - 2.395, esc 2 = esc (t - 2) = esc 1.14 = 1.101. Example 3-6. Find tan 4. Solution: As shown in Fig. 3-7(6), the reference number h is 4 — w = .86. Since the tangent is positive in the third quadrant, tan 4 = tan .86 = 1.162. Example 3-7. Find cos 5. Solution: Here, as shown in Fig. 3-7(c), h = 2tt — 5 = 1.28; and cos 5 = cos 1.28 = 0.2867. Example 3-8. Find cot 20. Solution: To locate the point P(20), we must proceed 20 units around the uni$ circle in a positive direction from (1, 0). The number of units in one complete revolution is 2w = 6.28, find we find that 20 = 3(6.28) + 1.16. Therefore, to locate P(20), we must proceed three times around the unit circle and then continue for 1.16 additional units in a counterclockwise direction. This means that t may be taken as 1.16. Since P(20) or P(1.16) lies in the first quadrant, ^ = t = 1.16. From the table, we have cot 20 = cot 1.16 = O.i 74 The Trigonometric Functions Sec. 3-4 Example 3-9. Find sin (-2). Solution: For <= - 2, it is shown in Fig. 3-8 that ti = | - t - (— 2)| = | — tt -f 2 1 = 1.14. Hence, t\ is the same as ^i in Example 3-5. Since sin t is negative in the third quadrant, we have sin (- 2) = - sin 2 = - sin 1.14 = - 0.9086. It should be noted that in Fig. 3-8 the points P(—t) and P(t) are located on oppo- site sides of the #-axis, and are joined by a line segment which is bisected perpendicu- larly by the axis. Hence the coordinates of the two points are numerically equal, but the ^/-coordinates have opposite signs. Therefore, by the definitions of the func- tions given in Section 3-2, it follows that Fig. 3-8. sin (— t) = —sin t, cos (-£) = cos t, tan (— t) = —tan t, esc (— £) = —esc t, sec (— t) = sec t, cot (— t) = — cot*. Hence, if t > 0, (3-18) (any function of (-*) | = |same function of t\. However, the algebraic sign of the function is changed for all functions except the cosine and the secant. Because cos t and sec t remain unchanged when t is replaced by its negative, these functions are called even functions. The remain- ing functions are called odd functions, since their values change sign when t is replaced by its negative. EXERCISE 3-3 1. Construct a figure, locating each of the following points. Show the point P(t) and its related number tu (Use tc = 3.14). a. P(l). b. P(3). c. P(10). d. P(- 5). e. P(- 4). f. P(3/2). 2. With the aid of Table I find each of the following values, using it = 3.14. a. sin 1.45. b. cos 3.5. c. sec 4.75. d. tan 5. e. esc (- 2.41). f. cot (- 4.50). g. sin 28. h. cos 60. i. tan ( - 30). . 3ir j. sec-£-« k. cot 13tt I. sin 22tt 3-5. POSITIVE AND NEGATIVE ANGLES AND STANDARD POSITION We have defined each of the six trigonometric functions as a relation between two sets of numbers, employing as the independent Sec. 3-5 The Trigonometric Functions 75 variable a real number t whose absolute value represents the length of an arc of a unit circle. Now we shall return to the tradi- tional viewpoint and consider trigonometric functions of angles. Although the stu- dent is probably famil- iar with the idea of angle from the study of geometry, we shall try to make the definition more precise. Let us select a point in a plane and draw the half- line or ray a emanating from O, as shown in Fig. 3-9. We shall call O the vertex of the ray. Finally, we let A be a point on the ray in its initial position. Now rotate the ray a about^O to some terminal position 6, so that the point A moves along the arc indicated by the curved arrow AB. The ray may be rotated in the counterclockwise sense, as in Fig. 3-9 (a) , or in the clockwise sense, as in view (6) . Moreover, it may be turned through one or more complete revolutions, as in view (c) . We shall speak of the position b as the terminal ray b. We have then an ordered pair of half -lines consisting of the initial ray a and the terminal ray 6. We can now define an angle as follows : An angle is a geometric figure consisting of two ordered rays emanating from a common vertex. With each angle is associated a number, called the measure of the angle, which indicates the sense and amount of rotation required to turn from the initial ray of the angle to the terminal ray. This rotation is usually represented graphically by a curved arrow. Its evaluation will be considered in Section 3-6. We may designate the angle in Fig. 3-9 as angle AOB; or we may use a Greek letter, such as 0, #, a, /3, or y, as the designation. The line OA is called the initial side of angle AOB, and OB is the terminal side. Counterclockwise rotation, as in Fig. 3-9 (a) oi 3-9 (c), gives rise tb a positive angle, while clockwise rotation, such as the one in Fig. 3-9(6), gives rise to a negative angle. Finally, we shall say that an angle is in standard position with respect to a rectangular coordinate system when its vertex is at the origin and its initial side coincides with the positive ataxia. See 76 The Trigonometric Functions Sec. 3-5 i \y V ■\ ^ I III IV *»x Fig. 3-10. Fig. 3-11. Fig. 3-10. When an angle is placed in standard position, the terminal side determines the quadrant to which an angle is said to belong. Thus, angle XOP in Fig. 3-10 is positive because it is generated in a counterclockwise direction, and is a second-quadrant angle because the terminal side OP lies in the second quadrant. We note that the definition of angle does not specify that the rotation should stop at the first arrival at the terminal side OP, Fig. 3-10. In fact, angles of any size may be generated, since any number of angles which end at the terminal side OP of a given angle may be obtained simply by adding a number of complete rotations, positive or negative, to the given angle. For example, the same terminal side may also be reached by rotation in the opposite direction. All angles which are in standard position and have the same terminal sides are called coterminal angles. In Fig. 3-11 the angle a is generated by rotation of OX counter- clockwise to the position OP. The angle /3, which is coterminal with a, is generated by adding to a one complete rotation of OX. The angle y is a negative angle, which is coterminal with a and is generated by rotating OX in the clockwise direction to the position OP. *»x 3-6. MEASUREMENT OF ANGLES The problem of measuring an angle is equivalent to that of finding the measure of the associated arc. One should, there- fore, apply the discussion of Section 3-1 and construct a unit circle as shown in Fig. 3-12. Let be an angle in standard position. Since the initial and terminal sides of the angle intersect the circle in the points P (0) Fig. 3-12. an( j P(t), respectively, the problem of measuring the angle reduces to that of measuring the appropriate $ec. 3—7 The Trigonometric Functions 77 arc length t Thus, the measure of the angle can be found in terms of a real number in any one of several ways, depending on the unit of measure chosen. We shall consider first the circular system, or natural system, of measuring angles, which is used almost exclusively in the calculus and its applications. Its fundamental unit is the radian. This unit may be defined as follows : A radian is the measure of an angle which, if placed at the center of a circle, intercepts an arc on the circumference equal in length to the radius of the circle. In Fig. 3-13 the angle AOB is 1 radian, and the length of the subtended arc AB is equal to the radius r. / \ arc =: radius *r If the circle selected for meas- uring a radian is a unit circle, we have an alternate definition of a radian. That is, a radian is an angle which intercepts a unit arc F IG , 3.43, on a unit circle. Another system of measuring angles is the sexagesimal system, or degree system, which is commonly used in ordinary calculations involving angles. The fundamental unit of this system is the degree. In Section 3-7 we shall study various relations between radians and degrees, and shall develop rules which allow us to convert from one system to the other. In discussing angles, we frequently use the term angle, in place of measure of an angle, and we rely on the context to make the meaning clear. Thus, when we say "0 = 2," we mean, "0 is an angle whose measure is 2 radians." The word radian is usually omitted when an angle is expressed in terms of radians. 3-7. THE RELATION BETWEEN RADIANS AND DEGREES Since an arc that is equal in length to the radius of a circle sub- tends an angle of one radian at the center, it follows that the whote circumference, which is 2n times the radius, subtends an angle of 2tt radians. Furthermore, the whole circumference subtends a central angle of 360°. Therefore, 27r radians = 360°, and ir radians = 180°. 78 The Trigonometric Functions Sec. 3-7 If the approximate value 3.1416 is used for 7r, 180° 180° or or Also, 1 radian = = t , 1g (approximately), 7T O.1410 1 radian = 57.29578° (approximately), 1 radian = 57°17 / 45 // (approximately). 1° = r— radians = 0.01745329 radians (approximately). loU In order to make the conversion to radians easier when the angle is expressed in degrees, minutes, and seconds, we give the following values : 1' = 0.00029089 radians, and 1" = 0.00000485 radians. Therefore, one of the following rules can be used to convert from degrees to radians or from radians to degrees : To convert from degrees to radians, multiply the number of degrees by ~ , or 0.0174533. loU To convert from radians to degrees, multiply the number of radians by — , or 57.29578. Note, however, that certain angles are commonly expressed in terms of tt radians, in order to avoid approximate values. For example, 180 = 7r radians, 45 = 7T/4 radians, 90° = 7r/2 radians, 30° = tt/6 radians. 3-8. ARC LENGTH AND AREA OF A SECTOR In Fig. 3-14 is shown a circle of radius r. In such a circle an angle at the center equal to one radian subtends an arc on the circumference equal to r. Similarly, by the definition of a radian, the number of units in the arc s intercepted by a central angle equal to 9 radians is given by the relationship .s r 5 = 1' Thus, (3-19) $ = rd, or Fig. 3-14. arc = (radius) • (central angle expressed in radians) . Sec. 3—8 The Trigonometric Functions 79 Now let A denote the area of the sector bounded by two radii and an arc of length s. If is the number of radians in the central angle of the sector, then the ratio of the area A of the sector to the area of the whole circle, or 7rr 2 , equals the ratio of the angle to the angle in the whole circle, or 2tt. That is, — = — > 7rr 2 27T or (3-20) A=^r 2 0. If the central angle of an arc or a sector is expressed in degrees, it must be re-expressed in radians before (3-19) or (3-20) can be applied. Example 3-10. Express 210° in terms of it radians. Solution: Since 1° = ^ radians, 210° = 210 • ~g = -g- radians. Thus, 210° = y radians. Example 3-11. Express 12°15'20" in radians. Solution: Multiply the decimal parts of a radian given in Section 3-7 for 1°, 1', and 1" by 12, 15, and 20, respectively. The results are as follows: 12° 0' 0" = .20943948 radians 15' 0" = .00436335 radians 20" = .00009700 radians 12° 15' 20" = .21389983 radians. Example 3-12. Express -^- radians in degrees. Solution: Since w radians = 180°, -^- radians = ~ (180°) = 150°. Example 3-13. Express 3.5 radians in degrees, minutes, and seconds. Solution: First, convert the radians to degrees, as follows: 3.5 radians = (3.5) (57.29578°) = 200.5352°. To find the number of minutes, multiply the decimal part of a degree, or 0.5352, by 60. Thus, 0.5352° = (60) (0.5352) minutes = 32.112'. To find thfc number of seconds, multiply the decimal part of a minute, or 0.112, by 60. The result is 0.112' = (60) (0.112) seconds = 6.72*. Hence, 3.5 radians = 200°32'6.72". 80 The Trigonometric Functions Sec. 3-8. Example 3-14. The radius of a circle is 5 inches. Find the length of the arc of the circle subtended by a central angle of 30°. Solution: Since 30° = -^ > the central angle 6 is -— • Also, r = 5. Therefore, by (3-19), «=ffl=5-£=5 (3.1416) = 2.618 inches, b o Example 3-15. In a circle of radius 6 inches, what is the area of a sector whose central angle is 60°? Solution: By (3-20), the area of the sector is ^r 2 0. Since 6 = 60° = ~ y 2i 3 1 7T A = o(36)"o~ = 6tt square inches. EXERCISE 3-4 In each problem from 1 to 25, express the given angle in radians. 1. 60°. 2. 45°. 3. 30°. 4. 10°. 5. 120°. 6. 150°. 7. 12°. 8. 90°. 9. 240°. 10. 330°. 11. 72°. 12. 20°. 13. 215°. 14. 196°. 15. 321°. 16. 283°. 17. 63°. 18. 30°10'. 19. 46°21'. 20. 236°37'. 21. 82°16'. 22. W2YIT. 23. 183°57'43". 24. 392°44'27". 25. 93°31'38' In each problem from 26 to 40, express the given angle in degrees. 26. tt/6. 27. tt/4. 28. ir/8. 29. 3tt/2. 30. 4tt/5. 31. tt/12. 32. 5ir/18. 33. 7tt/2. 34. 5x/3. 35. 3tt/20. 36. 3.7 rad. 37. 8.21 rad. 38. 0.34 rad. 39. 0.763 rad. 40. 0.8136 rad. In each problem from 41 to 56, draw the given angle in standard position and indicate its terminal side. 41. 30°. o 42. tt/4. 43. ir/3. 44. 90°. 45. 22^°- 46. 2t/3. 47. 170°. 48. 17x/18. 49. - t/2. 50. 630°. 51. 360°. 52. - 4tt. 53. 7tt/3. 54. 1000°. 55. 9x/4. 56. - 11tt/6. 57. In a circle of radius 4 feet, find the length of the arc intercepted by an angle of 7tt/Q radians. Find the angle in radians that intercepts a 5-foot arc. 58. A central angle in a circle of radius 15 inches intercepts an arc of 5 inches. Find the number of radians in the central angle. Express this angle in degrees and minutes, rounding off the result to the nearest minute. 59. A central angle of 62° 14' intercepts an arc of 16 inches on the circumference of a circle. Find the radius of the circle. 60. Find the area of a circular sector whose radius is 7 inches and whose central angle is a) 4 radians; b) 75°; c) 3 radians. 61. The area of a circular sector is 72 square inches. Find the angle if the radius is a) 6 inches; b) 9 inches; c) 5 feet. 62. The area of a circular sector is 126 square inches. Find the radius if the angle is a) 128°; 6) 1.6 radians; c) 30°. Sec 3-9 The Trigonometric Functions 81 3-9. TRIGONOMETRIC FUNCTIONS OF ANGLES Let be an angle in standard position, as shown in Fig. 3-15. With we can associate a real number t, which is the measure of the angle in radians. This concept is equivalent to our previous concept of t, when t was interpreted as the length of an arc laid off on the unit circle by starting at the point (1, 0) and terminating at *»x P(*,y) +x Fig. 3-15. Fig. 3-16. the point P{t). Such an association of the angle with the directed length t of an arc of a unit circle allows us to define the cosine and sine of 9 as cos = cos t and sin 9 = sin t. A similar procedure may be followed for the other functions of 9. Now consider Fig. 3-16, where we show an angle 9 in standard position and a unit circle. By the definition of 9, the terminal side of 9 intersects the unit circle at the point (cos 9, sin 9). This is, of course, the point designated previously as P(t). We now extend the terminal side of 9 to an arbitrary point P wi th coor dinates (x, y). The length of the radius vector OP is r = -\/x 2 + y 2 . If we drop perpendiculars from the points (cos 9, sin 0) and (x,y) to the #-axis, the right triangles thus constructed are similar. Therefore, x cos 9 t y __ sin 9 1 and - = r Hence, the coordinates of the point P(x,y) on the terminal side are x = r cos and y = r sin 9. Using these results with the definitions of the functions from Section 3-2, we caii express the values of the six functions in terms of x, y, and r. Thus, sin 9 = y/r } esc 9 = r/y, (3-21) cos 9 = x/r t sec 9 = r/x, tan 9 = y/x, cot 9 = x/y. 82 The Trigonometric Functions Sec. 3-10 3-10. TABLES OF NATURAL TRIGONOMETRIC FUNCTIONS OF ANGLES Tables of natural trigonometric functions are so labeled to dis- tinguish them from tables of the logarithms of these functions. Angles in Radians. In Section 3-4 Table I was used to find values of trigonometric functions of the type cos 2 or sin 27r/3. On the basis of the definitions of the functions of an angle given in Section 3-9, Table I may also be used to find the functions of angles meas- ured in radians. Example 3-16. Find the cosine of an angle of 1.43 radians. Solution: From Table I, cos 1.43 = 0.1403. Angles in Degrees. Table II at the end of this text contains the approximate values of the six functions of acute angles expressed in degrees and minutes. It is a four-place table of the functions of angles at intervals of 10 minutes. To find the value of a function of an angle between 0° and 45°, first locate the angle in one of the columns at the left, and then look for the value on the same line in the column headed by the name of the desired function. For an angle between 45 p and 90°, locate the angle in a column at the right, and then look for the value on the same line in the column with the name of the desired function at its foot. Table II should be referred to in working through these illustra- tive examples. Example 3-17. Find sin 32°40'. Solution: This angle is between 0° and 45°. Look in the left-hand column to find 32°40', and then go to the right to the column headed sin. There find 0.5398. Hence, sin 32°40' = 0.5398. Example 3-18. Find cos 56°20'. Solution: This angle is between 45° and 90°. So look in the right-hand column to find 56°20', noting that 56°20' is above 56°00', and then go to the left to the column with cos at its foot. Thus, cos 56°20' = 0.5544. The following examples illustrate the procedure for finding an angle corresponding to a given value of a function. Sec. 3-f-lO The Trigonometric Functions 83 Example 3-19. Given tan 6 = 4.511, find 0, assuming that 0° £ ^ 90°. Solution: Since tan is greater than 1, is greater than 45°. Therefore, search through the columns marked tan at the foot for the given number 4.511. The corresponding angle in the right-hand column is 77°30'. So 4.511 = tan 77°30', or = 77°30'. Example 3-20. Given cos = 0.8660, find 0, assuming that 0° ^ £ 90°. Solution: By looking through the columns with cos at either the head or the foot, find 0.8660. Since this value is in a column headed cos, use the left-hand column for the corresponding angle, which is 30°. Hence, = 30°. Interpolation. When either the given angle or the given value of a function is not printed in the table, we can find the desired value or angle by using a method of approximation known as interpola- tion. We assume that the change in the value of the function is directly proportional to the change in the angle. Although this assumption is not strictly valid, it gives values that are accurate enough for many practical purposes if we limit its use to small changes in the angle. The process of direct interpolation is used if the angle is given and we need to find the value of either an increasing function of the angle, such as the sine, or a decreasing function, such as the cosine. Inverse interpolation is used when the value of a trigo- nometric function is known and the angle is to be found. Example 3-21. Find sin 18°12'. Solution: This angle is not listed in the table, but it lies between 18 9 10' and 18°20'. From the table we find that sin 18°10' = 0.3118, and sin 18°20' = 0.3145. The desired value of sin 18°12' will then lie between 0.3118 and 0.3145. The tabular difference, that is, the difference between the two values listed in the table, is 0.0027. Also, the difference between the angles 18°10' and 18°20' is 10', while the angle 18°12' differs from 18°10' by 2'. Since the change in the angle from 18°10' to 18°12' is 2/10 of the change from 18°10' to 18°20', we assume that the corresponding change in the value of the sine will be (0,2) (0.0027) = 0.0005, and the amount to be added to 0.3118 is 0.0005. Hence, sin 18°12' = 0.3123. The accompanying diagrammatic arrangement presents this same operation in tabular form: , . A , J sin 18°10' = 0.3118 1 10 { J sin 18°12' = 0.3118 + x J \ 0.0027 sin 18°20' = 0.3145 84 The Trigonometric Functions Sec. 3-10 Since the angle 18°12' is 2/10 of the way from 18°10' to 18°20', the corresponding functional value will be 2/10 of the way from 0.3118 to 0.3145. Therefore, x 2_ 0.0027 10 ' x = (0.2) (0.0027) = 0.0005. This amount is to be added to 0.3118. Hence, the value of the function is 0.3123, and sin 18°12' = 0.3123. or Example 3-22. Find cos 73°48'. Solution: The process is similar to that in Example 3-21. However, since the cosine decreases as the angle increases, we subtract 8/10 of the tabular difference from cos 73°40'. We find the values of cos 73°40' and cos 73°50' in a column of the table labeled cos at the bottom. The work may be indicated as follows: cos 73°40 ; = 0.2812 8 { } x cos 73°48' = 0.2812 - x 10 cos 73°50' = 0.2784 0.0028 g^g = ft ' or x = (°* 8) (0 - 0028) = °- 0022 - Hence, the amount to be subtracted from 0.2812 is 0.0022, and cos 73°48' = 0.2790, The inverse process of finding the angle when the given value of a function is not printed in the table is performed in a similar fashion. Here, since we know the value of the function, we find the two values in the table nearest the given value, one less than it and one greater. Again making the assumption that small changes in the value of the function are proportional to small changes in the angle, we proceed as indicated in the following example. Example 3-23. Find if cot = 0.8780. Solution: This value of the cotangent is not in the table but lies between the entries 0.8796 and 0.8744. To these correspond, respectively, the angles 48°40' and 48°50'. We have, therefore, the following tabulation: f cot 48°40' = 0.8796 1 x \ 0.0016 10 { [ cot 48° (40 + x)' = 0.8780 J \ 0.0052 cot 48°50' = 0.8744 Hence, = 48°43 / . * 16 A 10 = 52 ,and * =3 ' Sec. 3-10 The Trigonometric Functions 85 EXERCISE 3-5 In each of the problems from 1 to 30, use Table II to find the value of the given function. Interpolate whenever necessary. 1. sin 36°20'. 2. cot 12S°40'. 3. sec 23°40'. 4. cos 96°50'. 5. sin 132°10'. 6. tan (- 2S°10'). 7. esc 223°30'. 8. sec 39°30'. 9. cot 283°50'. 10. sin 98°40'. 11. cos 75°C0'. 12. cot (- 133°30'). 13. sec (- 392°10'). 14. esc (- 416°20'). 15. tan G23°40 / . 16. tan 298°52'. 17. cot 55°43'. 18. esc (- 44°51'). 19. sin 57°32'. 20. cot 3°1G'. 21. sin (- 280°33'). 22. cot 28°01'. 23. tan 27°16'. 24. esc (- 245°29'). 25. cos (- 72°58'). 26. sin 312°37'. 27. tan 636°02'. 28. sin (- 16°47'). 29. esc 2S9°0G / . * 30. cos 12G°19'. In each of the problems from 31 to 60, use Table II to f.nd the values of between 0° and 360° which satisfy the given equation. Express the results to the nearest minute, interpolating whenever necessary. 31. tan = - 0.11CS. 34. cos0 =0.7951. 37. sin = 0.5783. 40. cos = - 0.4147. 43. tan = 8.345. 46. sin = 0.G702 49. ccs = 0.9503. 52. cot = - 1.381. 55. cot = 7.C00. 58. cot = 0.1340. In each of the problems irom Gl to 72, find the value of the given function. Interpolate whenever necessary. Take t as 3.14. 61. sin 0.93. (7. cot 2.46. 63. sec (- 1.24). 64. tan 8.71. C5. esc 9.43. 66. cot 0.G78. 3tt 4 ' 32. sin 6 = 0.30G2. 33. cot 6 = 1.091. 35. tan 6 = 0.0553. 36. sin 6 = 0.2419. 38. cot 6 = - 0.G494. 39. tan 6 = 1.511. 41. sin~0 = - 0.9959. 42. cot = 0.0437. 44. cot 6 = - 0.3121. 45. tan = 1.446. 47. tan 6 = 0.9043. 48. cot 6 = 2.398. 50. cos 6 = - 0.5090. 51. cos 6 = 0.S519. 53. cot 6 = 0.4230. 54. sin 6 = 0.2491. 56. tan = -0.1191. 57. ccs 6 = - 0.1323. 59. tan 6 = - LS.C0. 60. tan = 3.235. 67. tan 0.333. 68. cot ( - t) 69. cos 70. sin ( - ~) • 71. esc 0.968. 72. cot ( - 0.643). In each of the problems from 73 to 84, find the values of 6, in radians, between and 27T which satisfy the given equation. Use Table I and express the results to three decimal places, interpolating whenever necessary. 73. cos 6 = 0.9759. 74. sin 6 = 0.99G7. 75. tan = 2.066. 76. sec = 2.563. 77. tan = 0.9413. 78. sin = - 0.7360S 79. cos = 0.4010. 80. tan = l.G. 81. sin = 0.91. 82. cot 6 = 0.39. 83. cos = 0.84. 84. cot = 1.031. 4 The Laws of Exponents 4-1. POSITIVE INTEGRAL EXPONENTS When studying the progress of algebra up to the sixteenth century, one cannot help but be perplexed by either the total absence of symbolism or, when present, the lack of uniformity in its use. At first, unknown quantities were often represented by words. Later, symbols made from abbreviations and initial letters of these words were used to indicate mathematical concepts, such as number, power, and square. Descartes (1637) is generally credited with our present system of exponents. He introduced the Hindu-Arabic numerals as expo- nents, using the notations a, aa [sic] , a 3 , a 4 , etc. The writing of a repeated letter for the second power of the unknown continued for many years. Laws for positive integral exponents were introduced in Section 1-11, without proofs. We shall now establish these laws and extend them to apply also to zero, negative, and fractional exponents. We recall that if n is any positive integer, a n means the product of n factors each equal to a. In this notation, a is the base and n is the exponent or power. We shall proceed to establish the following laws for positive integral exponents. Law of Multiplication. If a is a real number, and if m and n are positive integers, (4-1) a m • a n = a m+n . Proof. Proof of this relationship follows from the definition of a n and the associative law for multiplication. Thus, a m = a • a a (tow factors), and Hence, a n = a * a a (ton factors). a m • a n = [a • a • • • • a(to m factors)] [a* a a(to n factors)] = a • a cr(to m + n factors) = a m+n For example, x s • x 5 = x 8 , and y k • y k+3 = y 2k + 3 . 86 Sec. 4—1 The Laws of Exponents 87 Law of Division. If a is a non-zero real number, and if m and n are positive integers such that m > n, then (4-2) Z- = a™~\ If a ¥= 0, and if n > m, then n m 1 (4-3) ^L = -J_ . v y a n a n - m Proof. Proofs of these relationships follow : If m > n, then m — n is positive. By (4-1) , a m-n . a n — a (m-n)+n — a m # Hence, dividing both sides by a n , we have a m a m ~ n = — • a n For example, -t = x 3 j and 7^ = 3 2 . If m < n, then n — m is positive. By (4-1), Divide both sides by a n ~ m to obtain a n a m = a n-m Now, dividing both sides by a n , we have a m ( a n \ / a n 1 1 a n \a n - m J / a n a n ~ m a n ~ m For example, z 4 1 , 3 5 1 5*"?' and 3* = 3*' Law for a Power of a Power. If a is a real number, and if m and n are positive integers, then (4-4) (a m ) n = a mn . Proof. This relationship can be easily proved as follows : By the associative laws for multiplication and addition, the law of multiplication expressed by (4-1) can be extended to three or more factors. Thus, a m • a n • a p = (a m • a n ) • a p = a m+n • a p A similar relationship can be written for any number of factors. That is, (a m ) (a*) • • • (a r ) = a m+ *+ •••+'. 88 The Laws of Exponents Sec. 4-1 We may now take m = p = • • • = r to get (a m ) (a m ) • • • (a w ) (to n factors) = a w + w+ •••+» = a»»n # Hence, (a w ) n = a mn . For example, (a; 2 ) 3 = x 6 , and (2 2 ** 1 ) 5 = 2 10A + 5 . Law for a Power of a Product. If a and 6 are real numbers, and if m and n are positive integers, then (4-5) (ab) n = a n b n . Proof. In proving this relationship, we make use of the associa- tive and commutative laws of multiplication. Thus, {ab) n = (ab) • (ab) (ab) (to n factors) = [a * a a(to n factors)] [b • b 6(to n factors)] = a n b n . For example, (- 2xV) 6 = (~ 2) 5 s 1 V 6 = - 32z 1 V 5 . Law for a Power of a Quotient. If a and b are real numbers, if 6 ¥" 0, and if n is a positive integer, then Proof. By applying the law for multiplying fractions, we have W =6*6 ^ (ton factors) =^. For example, (3xy k \ 2 _3 2 x 2 y 2k _ 9s 2 y 2 * V z 2 / " z 4 ~ z* ' An exponent affects only that quantity to which it is attached. Thus, ~5x(y 3 ) 2 = — 5xy Q , whereas (— 5xy 3 ) 2 = 25x 2 y Q . So far we have defined a n only when n is a positive integer. We shall now introduce zero, negative integer* and rational powers in such a way that they will obey the same laws which were proved for positive integral exponents. 4-2. MEANING OF a We shall define the zero exponent by the equation (4-7) a° = 1 (a^ 0). A few illustrations are : aP = l, 7° = 1, 5(2zt/*)o = 5, (a» - 6*)° = 1, (^)° = 1. Sec. 4-2 The Laws of Exponents 89 If a in (4-2) is not zero and m = n, we get a m — = a m ~ n = a . a n In this case, the quotient on the left equals 1, while the value of the term on the right is a . Since a = 1, by definition, the law of division holds for n = m, as well as for m> n and n> m. The student should note that (4-7) gives the only possible defini- tion of a if the law expressed by (4-2) and (4-3) is to hold for the zero exponent, as can be seen from the foregoing discussion. We shall show that the definition a = 1 is consistent with the five laws of exponents in Section 4-1 ; that is, we shall show that these laws also hold when any exponent is zero. In the following explanations, where a quantity occurs in a denominator, we assume that it is not zero. Also, the exponents are assumed to be non- negative integers. Let us, for sake of discussion, suppose that n = in (4-1), that 1S > 1X1 a m . a n = a m+n m Then we have n m 1 a m • a n = a m • a = a m • 1 = a m , or Hence, the law of multiplication holds when n = 0. A similar procedure will verify the law if m = 0. Now let us suppose that n = in (4-2), that is, in a m — = a m ~ n . Then , Qin Qin Qta — = — = —- = a m , and a m ~ n = a m ~° = a w . a n a 1 ' Hence, (4-2) holds when n = 0. If m = in (4-3) , we have £!! - ^ - J_ , d 1 1 = 1 t a n ~" a n ~" a n ' a n ~ m ~~ a n_0 ~" a n It is clear that in (4-2) m cannot be zero, and in (4-3) n cannot be zero. Therefore, the law of division holds. Suppose that n = in (4-4) , that is, in (a m ) n = a mn . Then (a m ) n =i (a w )° = 1, and a mn = a w, ° = a = 1. If m = in (4-4) , we have (a™) n = (a°) n = l n = 1, and a mn = a°' n = a = 1. Hence, the law for a power of a power holds. 90 The Laws of Exponents Sec. 4-2 Now consider (4-5) , which is (db) n = a*b n . If n = 0, (ab)» = (ab)° = 1, and a»b n = a°b° = 1 • 1 = 1. Finally, we let n = in (4-6), that is, in (ay_ <r W ~ b n ' Then , ffi> - - o . a » a o j y = w =i ' and t«=¥ = i =i - The demonstrations just given prove that the five laws of expo- nents, originally stated for positive integral powers, are true for all non-negative integral powers, and that the law of division is true even when the exponents are equal. 4-3. NEGATIVE EXPONENTS In order to extend the meaning of exponents to negative integers, we define ar n by the following relationship : (4-8) or- = ± {a* 0), where n is a positive integer. Several illustrations are: 5 - 2= ^=4' 1^ = 103 = 1000, (a - to) ^ = (T^Eo 5 ' and @ 1= i = i' y As in Section 4-2, we shall show that our definition is consistent with the five laws of exponents. Let us first note that (4-8) is true even if n = or if n is a nega- tive integer. If n = 0, then a -n = a o = ! = 1 J_ = 1 . 1 o° a n If n = — p, where p is a positive integer, then a~ n = a p = 1 /— ~~ er p ~~ a n # We shall use this result in the proofs that follow. In order to extend (4-1), let m be a non-negative integer, and let n be a negative integer, say, w = —p. In this case it is also assumed that a ¥> 0. Then 1 a m a m • a n = a m • a~ p = a m • — = — • a p a p Sec, 4—3 The Laws of Exponents 91 If m ^ p, we have, by (4-2) for non-negative exponents, d m a m • a n = — = a m ~ p = a m+( ~ p) = a m+n . a p If m<p, we have, by (4-3) for non-negative exponents, o m 1 a m • a w = — = = a~ (p ~ m) = a w ~ p = a m+n . A similar demonstration establishes (4-1) in case n ^ and m < 0, or in case ra < 0, and n < 0. Proof of the extension of (4-2) rests on the validity of the law of multiplication just established. If a¥=0, and if m and n are integers (positive, negative, or zero), then a m 1 — = a m — = a m • ar n = a w_n . a n a n Although (4-3) is now an immediate consequence, it is not really needed, in view of the general validity of (4-8). The demonstra- tion just given allows the law of division to be stated as a single relationship as follows : (4-9) *— = a m ~ n (a^O). Thus, a single law applies, regardless of whether m > n, n > m, or m = n, where m and n are arbitrary integers. Now consider the law for a power of a power. In (4-4) let n — —p, where p^O, while m ^ 0. Then K a ) - w ) - (am)p amp Also, 1 a mn = a m (~ p ^ = a —,np = • Hence, (4-4) holds in this case. If m ~ —p y where p ^ 0, while n S 0, we have / 1 \ w 1 1 (a™)" = (ar p ) n = ( — ) = — i and a mn = a ( ~ p)n = a~ pn = — • v J K J \a p / a np a np Again (4-4) holds. If both m and n are negative, a similar proce- dure is used, and the extension of (4-4) holds. To extend the law for a power of a product, let n = —p in (4-5), where p ^ 0. Then (a&) n = (a6)~ p = y-^- = —r- > and a n b n = a- p 6~ p = — . — = — -$ v ' v ' (a6) p a p b p a p b p a p b p So (4-5) is verified for negative integral values of n. To verify the exltended law for a power of a quotient, assume that n = — p in (4-6), where p = 0. Then W "" W "" /ay ~ /6* " 6 p , a n cr* 6 P -— > ana tz =r 7~r = -z a? b n * b~ p o? 92 The Laws of Exponents Sec. 4-3 Hence, (4-6) holds for negative integral values of n. Thus, the laws of exponents hold for positive integral exponents, zero exponents, and negative integral exponents. In Section 4-5 we shall consider the case of fractional exponents. From the general validity of (4-8), it follows immediately that a factor of the numerator or the denominator of a fraction can be moved from the numerator to the denominator, or vice versa, pro- vided only that we change the sign of its exponent. For example, a*b~° = tq > and b 3 yz~ 2 y SCIENTIFIC NOTATION We are now in a position to introduce certain simplifications when operating with very large or very small numbers, as are customarily used in scientific writing. Any positive number that is greater than 10 or less than 1 may be written compactly by expressing it in standard form, that is, by writing it as a number that lies between 1 and 10 multiplied by a suitable positive or nega- tive integral power of 10. Thus, 27,000 would be written 2.7 • 10 4 . Similarly, 0.00031 would be 3.1 • 10~ 4 . Example 4-1. The speed of light is 186,000 miles per second. Express this number in scientific notation. Solution: The given number 186,000 may be written as 1.86 • 10 5 . Example 4-2. The rest mass of an electron is 9.11 • 10~ 28 grams. How many zeros would be required between the decimal point and the first non-zero digit, 9, if the number were written in decimal notation? Solution: The exponent — 28 means that we would have to move the decimal point 28 places to the left from its present position. We would thus have to place 27 zeros to the left of the 9. Example 4-3. If the sun is 9.3 • 10 7 miles from the earth, how long does it take light to reach the earth from the sun? Solution: As given in Example 4-1, the speed of light is 1.86* 10 5 miles per 9 3 • 10 7 second. Therefore, the required time is z ■ -' ; ^ = 5 • 10 2 = 500 seconds = 8 minutes 20 seconds. 4-5. RATIONAL EXPONENTS We shall now extend the meaning of exponents from integers to rational numbers. Here again we shall make the extension in such Sec. 4-5 The Laws of Exponents 93 a way that the laws for positive integral exponents will be preserved. Suppose that a is a real number and that n is a positive integer. Let us assume that a 1/n has meaning and that (4-4) applies. Then it would be true that (4-10) {a l ' n ) n = a^ ,n)n = a 1 = a. This says that the nth power of a Vn would have to be a, or in other words that a 1/n would be what is called an nth root of a. For example, (4-10) would yield (a 1 ' 2 ) 2 = a, and (a 1 / 3 ) 3 = a. Real nth Roots of a. Before defining a 1/n , let us examine the sit- uation with respect to the existence of nth roots of a given number a. The following results may be proved with the help of the theory of equations. Case I. If n is an even integer and a is a positive real number, there are two real numbers that satisfy the equation r n = a. One of these is the positive nth root of a, which is denoted by tya. The other is the negative nth root of a, which is denoted by — tya. We may also denote these two numbers together by zt^J/a. Case II. If n is an even integer and a is a negative real number, no real nth roots exist, since no even power of a real number can be negative. Case III. If n is an odd integer and a is a positive real number, there is one real (positive) value of r such that r n = a. In other words, if n is odd, there is a real positive nth root, which is denoted by tya. Case IV. If n is an odd integer and a is a negative real number, there exists one real (negative) value of r such that r n — a. That is, if n is odd, there is a real negative nth root, which is denoted by Case V. If n is any positive integer and a is zero, there is only one real nth root, and this root is zero. Thus, the definition of an nth root of a is valid under all condi- tions except when a is negative and n is even. In this situation, no real nth roots exist. (However, the introduction of complex num- bers in Chapter 11 will allow us to eliminate this exception.) We are now ready for the following definition. * Definition. If a is a non-negative real number and n is a positive integer, a Vn designates the non-negative nth root of a, or tya. If a is negative and n is an odd positive integer, then a 1/n designates the real nth root of a, or ^/"a. When a is negative and n is even, a 1/n is undefined: 94 The Laws of Exponents Sec. 4-5 Meaning of a m/n . Let a be a given real number, n a positive integer, and m an integer. If a m/n has meaning, and if (4-4) holds for fractional powers, then a m/n = a (1/n)TO = (a l/n ) m . Under these assumptions, then, a m/n would be the mth power of a 1/n . It is natural to state the following definition. Definition* If n is a positive integer, if m is any integer such that the fraction m/n is in lowest terms, and if a is a real number which is assumed to be non-negative when n is even, then a m/n designates the mth power of a 1/n , that is, the mth power of tfa. Hence, (4-11) a mln = (a 1/n ) m . If the fraction m/n is not in its lowest terms, it is first reduced to lowest terms, and (4-11) is then applied. When a is given, the value of a m/n depends only on the value of the fractional exponent, not on the particular values of m and n. Thus, 2 4/2 = 2 2 = 4, 2** = 2 3 ' 4 , and (- 2) 2 ' 6 = (- 2) 1 ' 3 = JT^2. In the last example it would be incorrect to apply (4-11) directly, since (— 2) 1/6 has no meaning. It may be shown that, if a is positive, (4-12) a mfn = *s/cr. The proof is omitted. We shall also omit the details of the procedure for showing that the five laws of exponents hold for rational exponents and non- negative bases. The reader is cautioned against using the laws for negative bases, since some fail under certain conditions. To summarize the results now established, we restate the laws of exponents here for easy reference. It is assumed that a and b are non-negative real numbers, and that m and n are rational numbers. Furthermore, if either a or 6 appears in a denominator or raised to a negative or zero power, it is assumed to be different from zero. Law of multiplication: a m * a n = a m+n . d m Law of division: — = a m ~ n . a n Law for a power of a power: (a m ) n = a wn . Law for a power of a product: (ab) n = a n b n . (a\ n a n h) ~ fe~ # Law for reciprocal: a~ n = ~ • Zero power: a = 1. Sec. 4-5 The Laws of Exponents 95 Note that the radical notation can be replaced by the simpler and much more convenient exponential form. Everything that can be done with the radical notation in the simplification of roots of numbers and in operations involving roots can be done much more naturally by means of the exponential notation. A few illustrations of the meanings and uses of exponential forms follow : x 1 ' 2 = y/i, if x ^ 0; (- 27) 1 ' 3 = ^/^27 = - 3; - (32) 1 ' 5 = - S/32 = - 2; - ^fi = - a 2 ; 32 2 ' 5 = (^32) 2 = 2 2 = 4; - (16a; 4 )- 1 / 2 = " * "~ * "" l (16a; 4 ) 1 / 2 y/Tfat 4x 2 The following examples illustrate the applications of the laws of exponents to the solution of problems involving radicals. Example 4-4. Compute the value of \/2 • y/% and write the result in expo- nential form. Solution: Using exponential notation and the laws of exponents, we have 2i/3 . 21/4 = 24/u? . 2 3 ' 12 = 2 4/12+3/12 = 2 7/12 . Example 4-5. Remove all possible factors from the radical ^162x 4 2/ 2 . Solution: We may proceed as follows: ^162z 4 z/ 2 = (2 • 3 4 *V) 1/8 = 2 1 ' 3 • 3 4 ' 3 .r 4 ' 3 ?/ 2 ' 3 = 3x • 2 1 ' 3 3 1/3 z 1/3 2/ 2 ' 3 = 3x VOxy*. Example 4-6. Use the laws of exponents to express y/x • *tyy by using only one radical. Solution: Changing to fractional exponents, we have y/x* y/y = x ll2 y 113 = z 3 'V /6 = (» 8 2/ 8 ) 1/a = \/» a V 2 - Example 4-7. Rationalize the denominator in the fraction ~^== • Solution: To write an equivalent fraction in which no radical appears in the denominator, we proceed as follows: __ \_ 1 _ 1 s 2/5 _ s 2/5 y/&_ # ^/x z ~~ X 315 ~~ X 315 * x 2 ' 5 ~ x ~~ x Example 4-8. Ratiorialize the denominator of the fraction -= • 5 - V3 Solution: We use the relationship (a +b) (a — b) = a 2 — b 2 to remove the radical from the denominator. Thus, 1 1 . 8 + \/ 8 = 5 + V 3 - 5 ± V 3 - 5 - V 3 5.- V3 5 + V3 25-3 22 96 The Laws of Exponents Sec. 4-5 x 2 <\A 2 - x 2 + ■ *Aj2 #2 Example 4-9. Change _ \ to a simple fraction. Solution: Write the expression in exponential form, as follows: <* -«■>"' + («..!%.>. a 2 — a; 2 Then, multiplying the main numerator and the main denominator by (a 2 - x 2 ) 112 , we have „ 9 , 9 9 a 2 - .r 2 + x 2 a 2 (a 2 - x 2 ) 3 ' 2 (a 2 - x 2 ) 3f2 Example 4-10. Express (x 2 + a 2 ) 3 ' 2 + 3z 2 (x 2 + a 2 ) 1 ' 2 in a factored form. Solution: Rewrite the expression as {x 2 + a 2 ) 112 (x 2 + a 2 ) + Sx 2 (x 2 + a 2 ) 112 . Removing the common factor (x 2 + a 2 ) 112 , we obtain (x 2 + a 2 ) 112 [x 2 + a 2 -j- 3a; 2 ] = (x 2 + a 2 ) 1 / 2 (4a; 2 + a 2 ). EXERCISE 4-1 In each of the problems from 1 to 20, perform the indicated operations and eliminate all zero and negative exponents. 1. 3z°y. 2. ar 1 ' 2 . 3. (|)~ 3 - 4. 10- 2 . 5. j~zr 6. (|)°- 7. (9*)i/». 8. faV«-». 9. (»•)-»'*. 10. fo.V'i, • 11- ^f^ • 12. 3s W • ±x" 2 yz\ (a? l/4 2/) 4 a 2 t/" 9 y ^ 13. (a;V)*(^"V 4 ) 2 . 14. (& l 'V a ) 4 (s a y" 1/2 )°- 15- (* 1/8 H-y l/a ) 2 . 16. ar* + y x . 17. (a; + y)" 1 . 18. (x + y)' lf3 (x + 2/) 1/3 . 19 ! - *" 1/2 . 20 * 1/2 + 3" 1/2 , 1 + ar 1 ' 2 ' * a: 1 ' 2 -a;- 1 / 2 ' Write each of the following expressions in exponential form. Remove all possible factors from the radical and, wherever necessary, rationalize the denominator. 21. V§6. 22. ^~=~54. 23. ^2l. 24. y/xy*. 25. tfflyi. 26. </Fi. 27. v/l2r=^. 28. ^"^. 29. ^(-64)°y-». 30. Wx°y-»)*. 31. V^ly" 2 )" 1 . 32. vV + b 2 )- 1 . 33. V3 • ^14. 34. -#? • V5- 35. V^ v^. 36. #ai&5 - 4 /8^. 37. — 5_ = . 38. —J 39. — 2_ . 40. X 3 - V2 V6 + 2 1 - V5 V3 - V2 41. i_. « B 5±3£ I =». tt l+ ^^ u.*£=J^. 3-ya: 2 «-9 3-vz 2 -9 z-y/z*-y* Vz+V^+l Sec. 4-7 The Laws of Exponents 97 Vl - X 2 + —~= x(l - x) Aa y/\ - x* 45. V2x - x 2 + * v . .JL . 46, 1 + Vl + * 2 - y/2x - x 2 \ - x 2 x 2 47. __ — -^^jQ+^L . 48. ( x a + 1)3/2 + X 2( X 2 + 1)1/2. (1 + VI + * 2 ) 2 49. (2 - x 2 ) 5/2 + .r 2 (2 - x 2 )* 12 . 50. (a* - 3) I/2 - z 2 (z 2 - 3)" 1 ' 2 . 4-6. THE FACTORIAL SYMBOL The product of all positive integers from 1 to n inclusive is called "n factorial" or "factorial n" and is represented by either of the symbols n ! or An. Thus, if n is a positive integer, n! = 1 -2-3 (n - 1) • n. For example, 3! = 1-2-3 = 6; 5! = 1-2- 3-4- 5 = 120; 6! = 5!- 6; j = 7!; r! = [(r - l)!]r. 4-7. THE BINOMIAL THEOREM The statement known as the binomial theorem enables us to express any power of a binomial as a sum of terms without per- forming the multiplications. By actually performing the indicated multiplications, we find that (a + 6) 2 = a 2 + 2ab + 6 2 , (a + b) z = a 3 + 3a 2 b + Sab 2 + fc 3 , (a + b) 4 = a 4 + 4a 3 6 + 6a 2 b 2 + 4afe 3 + ft 4 . These formulas may be rewritten in the following manner, so as to suggest a general rule 1 : (a + b) 2 = a 2 + \ab + < ~b 2 i (a + 6) 3 = a 3 +\a% + f-fjafc* + f-ff^ 3 , (a + ^ = a* + fa 3 6 + *f§«V + f^fafc 3 + f^^*. Applying this suggested rule to (a + 6) 5 , we obtain (a + &)« = as +^ a«6 + £l| a»6» + ^|t| "^ , 5-4-3-2 fc4 , 5 « 4 » 3 • 2 • 1 ,,, + 1.2.3.4 qb4+ l-2.3.4.5 b - 1 The justification for writing the expressions on the right in this form will be found in Chapter 17, where the binomial coefficients are 'given in terms of the combination formulas. 98 The Laws of Exponents Sec. 4-7 Upon simplification of coefficients, we get (<x + b) 5 = a 5 + 5a 4 b + 10a 3 6 2 + 10a 2 6 3 -I- 5a6 4 + 6 5 , which is the same result as that obtained by multiplying (a + &) 4 by (a+6). Each of the expressions on the left is of the form (a + b) n , in which the exponents 2, 3, and 4 of (a + b) are special values of n. If we let n denote the exponent of (a + b) in each of the expres- sions on the left, we note that the expansion of (a + b) n contains n + 1 terms with the following properties : 1. In any term the sum of the exponents of a and b is n. Also, the first term is a n and the last term is b n . 2. The exponent of a decreases by 1, and the exponent of 6 increases by 1, from term to term. 3. The denominator of the coefficient in each term is the fac- torial of the exponent of b in that term. 4. The numerator of the coefficient in each term has the same number of factors as the denominator. Specifically, wherever 1 appears in the denominator, write n directly above it in the numerator ; wherever 2 appears in the denominator, write n — 1 directly above it in the numerator; and so on. Thus, in (a + b) 5 , the number above 1 is 5, and the number above 2 is 4. Assuming that these properties hold for all positive integral values of n, we have (4-13) (a + b) n = a» + j a^b + nf * ~ 1} a- 2 6 2 , n(n — 1) (n — 2) n „, a . , , „ + — — 1.2*3 + * " ' + This result is the binomial formula. So far we have verified this formula only for n = 2, 3, 4, and 5. In Chapter 16, we shall prove the binomial theorem, which states that the formula is true for all positive integral values of n. The following example shows the procedure for the expansion of (a + 6)\ Example 4-11. Expand (x - 2?/) 6 by the binomial theorem. Solution: The required expansion will be obtained by letting a = x, b = — 2y, and n = 6. We begin by setting up the following pattern of n + 1 terms : x 6 + s*( - 2y) + x*{ - 2y) 2 + x 3 ( - 2?/) 3 + x*( - 2yY + x( - 2yY + ( - 2y)\ Sec. 4-8 The Laws of Exponents 99 The exponents of b in the second, third, fourth, and fifth terms are 1, 2, 3, 4, and 5, respectively. Hence, remembering that the last term is b n , we may fill in the numerators and denominators of the coefficients as follows: *• + j *•(- 2y) + 5l| *«(- 2»)» + |f|f| «•(- 2»)« + \\\'X.\ **{- W By simplifying, we obtain the following result: (x - 22/) 6 = z 6 - 12x 5 2/ + 60s 4 i/ 2 - 160z 3 2/ 3 + 240x 2 */ 4 - 192xy* + 64y 6 . 4-8. GENERAL TERM IN THE BINOMIAL EXPANSION If we wish to write any particular term of the expansion of (a + b) n without considering any of the other terms, a study of the binomial formula in (4-13) Section 4-7 will reveal the' following facts : In every term the exponent of 6 is one less than the number of the term. Thus, in the (r+ l)th term, the exponent of b is r. (The expression for a particular term is simplified slightly if the number of that term is called r + 1, rather than r.) The sum of the exponents of a and 6 is n in each term. For the (r 4- l)th term the exponent of a is n — r. The denominator of the coefficient in the (r + l)th term is r!, since it is the factorial of the exponent of b. The numerator of the coefficient has the same number of fac- tors as the denominator. In the (r + l)th term, it is the product n(n — 1) (n — 2) • • • (n — r + 1). We obtain, then, for the (r + l)th term of the expansion (« + &)», n(n - 1) (n - 2) ■ ■ ■ (n - r + 1) alMftr r! Example 4-12. Find jihe sixth term of (3x - y 2 )*. Solution: Here a = 3#, b = - y 2 , and n = 8. Since r -f 1 = 6, the exponent of b is r = 5. Hence, the exponent of a is n - 5 = 3. Therefore, the sixth term is 8-7-6-S-4 1*2<3*4*5 (3x)H-V 2 ) 5 = -15123V . 1 00 The Laws of Exponents Sec. 4-8 EXERCISE 4-2 In each of the problems from 1 to 16, reduce the given fraction to lowest terms. 1. 5. 9. (91) (31) , 8! (3-4)!, 3(4!) n! (n-2)!' 2. 6. 10. 10! (3!) (7!) (3-4)! (3!) (4!) n! 13. 15. (n - r) ! (» + l)!(n -1)! (n!)» n!(n + l)l 3. 7. 11. (10(3!) , (2!) (2!) 3! +4! (3!) (4!) * (n + 1)! (n-1)!' [(n + 1)!] 2 n\(n -2)! n\(n -2)! 4. 8. 12. (7!) (8!) , (5!) (6!) nl (n-1)!" n! 3!(»-3)!' 17. Show that (n -l)!(n+2)l n(n - 1) (n - 2) 14. 16. [(n-l)l]» (n-r + 1) _ n! r! r! (w — r)! In each of the problems from 18 to 32, expand the given expression by the binomial theorem. (Hint: In problems 29 thru 32, first consider the first two terms in parentheses as a single quantity). 18. (x + y): 19. (* - 1) T . 20. (a - 26)*. 23.(yx+-tY - fx 2>6 21. (2a* - 362)'. 25.P-4)\ \y x 2 / 22. (Vx + y/y)\ 23. (V* + ^~=J . 24. (| - ^f . 26. (- * + y~ 2 Y. 27. (x» - y 2 )\ 28. (^ + ~^) 5 . 29. (» + 2/ + z) 2 . 30. (* + 2y + z) 2 . 31. (a; 2 + x + l) 4 . 32. (a 2 - a - l) 3 . In each of the problems from 33 to 42, find the indicated term. 33. (1 - x)*, 8th term. 34. (m + n)", 10th term. 35. (a + 26) 12 , 5th term. 36. (a - &)» 3rd term. - — J , 4th term. 38. {x - y) 10 , term involving t/ 4 . ^ J , term involving — • y *" * y 41. (y/x - y/y) 12 , middle term. 42. (2x - #) 7 , middle terms. 5 Logarithms 5-1. DEFINITION OF A LOGARITHM We shall assume here that the laws of exponents stated in Chapter 4 for rational exponents ark valid also for irrational exponents. The definition of a base raised to an irrational power is beyond the scope of this book. However, let us make the assump- tion that, if b and x are real numbers, with b positive, a corre- sponding number designated by b x exists. Without giving an explicit rule for computing b x , let us assume that all laws of expo- nents established in Chapter ~4 are valid generally for real powers. Finally, let us assume that, corresponding to any two positive real numbers & and n, where b ¥=■ 1, there exists a unique real number x, such that n = b x . We can then give the following definition. Definition. If n = b x , where 6 is a positive real number different from 1, then x is called the logarithm of n to the base 6. We write x = log & n. The following table shows both forms of several equiva- lent statements. Exponential Form 2 3 =8 . 41/2 = 2 6 "81 5° = 1 Logarithmic Form log 2 8 = 3 log 4 2 = \ log 3 gf = - 4 log 5 1=0 We shall restrict n to positive numbers, since negative numbers do not have real logarithms. ' For any positive base b, we have 6° = 1 and 6 1 = 6. Hence, it fol- lows from the definition of a logarithm that log b 1 = and log& b = f l. Another valuable, relationship results from combining the two equations n = b x and x = log& n. Replacing x in the first equation by its value from the second equation, we have n = &log„ % m For example, 2 10 ** 8 = 8, and 10 lo *i<> w = x. 101 1 02 Logarithms Sec. 5-1 Example 5-1. Find n, if logs n = 2. • Solution: Write the given equation in exponential form, as follows: „ w = 32 - Hence, n = 9. Example 5-2. Find the base 6, if log* 4 = 2/3. Solution: In exponential form, the given equation is b 213 =4. Raise both sides to the 3/2 power and recall that b > 0. Then (£2/3)3/2 = ^ = 43/2 # Therefore, 6=8. Example 5-3. Find x f if logi/ 8 32 = x. Solution: Writing the equation in exponential form, we have Express 1/8 and 32 as powers of 2, and get 1/8 = 1/2 3 = 2" 3 , and 32 = 2 5 . Hence, (2-3)* = 2 5 , or 2- 8 * = 2 5 . From this, we have - 3z = 5, and a; = - 5/3. Therefore, logi/g 32 = - 5/3. 5-2. LAWS OF LOGARITHMS Since a logarithm is an exponent with respect to a given base, the rules for operating with logarithms are the same as the laws of exponents. These laws, expressed in terms of logarithms, have the following form. Law I. The logarithm of a product equals the sum of the loga- rithms of its factors. The logarithmic form is (5-1) log& (m • n) = log 6 m + log6 n. Proof: To prove this equation, let x = log& m and y = log& n. Then m = b x and n = b y . Multiplying, we have mn =6^. Hence, . log 6 (mn) = x + y = log 6 m + log& n. Law II. The logarithm of a quotient equals the logarithm of the dividend minus the logarithm of the divisor. The logarithmic form is (6-2) log 6 (~) = log 6 m - log 6 n. Sec. 5-2 Logarithms 1 03 Proof: The proof follows : Let x = logb m and y = log& n. Then ,„ , ,., m = b* and n = b y . Dividing, we have _ - = b<~y. n Hence, /m . iogft ^— y = a - 2/ = log b m ~ l°g& n * Law III. The logarithm of a power of a number equals the expo- nent times the logarithm of the number ; that is, (5-3) log*, (n fc ) = k log 6 n. Proof: The first step in the proof is to let x = logb n. Then . n = o z . Raise both sides to the fcth power and obtain This relationship, when written in logarithmic form, becomes log 6 (n k ) = to. Replacing # by its value, we have log& (n k ) = & log6 n. The student should note carefully the difference between log& (n k ) and (log&n) fc . Law IV. The logarithm of a root of a number equals the loga- rithm of the number divided by the index of the root ; that is, (5-4) log 6 y/n = t log b n. Proof: This equation follows as a corollary of law III. By the definition of a fractional exponent, we have ^/n = n 1/k . Hence, by law III, log 6 y/n = log& (n llk ) = ^log* n. f a/51 Example 5-4. Express log2 — ^- as a linear combination of logarithms. iSerftiton: log 2 ^~P = log 2 VSl - log 2 3 4 = log 2 (3 • 17V'» - log 2 3 4 = log 2 3 1 '? + log 2 17 1 ' 2 - log 2 3 4 . 104 Logarithms Sec. 5-2 Example 5-5. Express 2 logio 3 - x 1°8 10 x + 1°S 10 2/ as a single logarithm. Solution: 2 logio 3 - ~ ^°Sio a; + logio y = logio 3 2 - logio x 112 + logio 2/ = logi (3 2 -y) - logio x 1 ' 2 , 3 2 2/ , 9y Example 5-6. Transform the equation logo # + y = log a sin x into an equation free of logarithms. Solution: By transposing, we get sin x 2/ = loga sin * - logo x = logo -— • Change to the following exponential form: sin x w = • EXERCISE 5-1 In each of the problems from 1 to 12, write the equation in logarithmic form. 1. 2 3 = 8. 2. 2« = 64. 3. 3 4 = 81. 4. 10° = 1. 5. 10 3 = 1000. 6. 10- 3 = 0.001. 7. 256 1 ' 8 = 2. 8. 216 1 ' 3 = 6. 9. 100 - 6 = 10. 10. y=ep. 11. 10 v = x. 12. 10 lo « v = x . In each of the problems from 13 to 21, write the equation in exponential form. 13. logs 64 = 2. 14. logs 125 = 3. 15. log 2 ^ = - 6. 16. log* ^ = - 4. 17. log 7 343 = 3. 18. log 9 729 = 3. two 19. logio 10,000 = 4. 20. logio 0.0001 = - 4. 21. log 4 8 = 3/2. In each of the problems from 22 to 33, find the indicated value of x. 22. logo 3 = x. 23. log 2 64 = x. 24. log 4 x = 0. 25. log* 4=2. 26. logo.s x = - 1. 27. log 3 x = 1. 28. log, 81 = 4. 29. log, 100 = - 2. 30. log, ^ = 5. 7 31. logo 243 = x. 32. log 64 x = - ^ • 33. log, a = 2. In each of the problems from 34 to 39, use the laws of logarithms to write the expression as a single logarithm. 34. log& 2-3 log* 5 + log 6 7. 35. log& 4 + log* w - log* 3+3 log* r. 19 1 2^ 36. ± log* 7 + 1 log* 4 + i log* 3. 37. - 5 log* 23 + 12 log* y • 38. 3 log* 2 + log* 13-2 log* 5. Sec. 5-4 Logarithms 105 39. ^ log* {u - Vu 2 - a 2 ) - ^ logb (u + s/u 2 - a 2 ) + log& a. 40. Find the logarithm to the base b of the area of a circle in terms of the logarithms of 7r and the radius. 41. The time T for a pendulum of length I to make one oscillation is T = ir *y - > where g is a constant representing the acceleration due to gravity, a) Find logb T in terms of the logarithms of tt, I, and g. b) Find log& I in terms of the logarithms of w, 1\ and g. 42. The area of a triangle with sides of length a, b y and c is given by the formula K = V$(-s — a) (s - 6) (s — c), where s is the semi-perimeter ^ (a -f 6 + c). Find log& X in terms of the logarithms of combinations of a, 6, and c. 43. The positive geometric mean G of n positive numbers Xi, x 2 , ■ • • . x n is defined by the relationship , n log 6 Ji + log 6 x 2 + • • • -f log& z n log* G = -s Show that G = v^xi.^ • • • x„. 5-3. SYSTEMS OF LOGARITHMS As we mentioned in Section 5-1, any positive number b different from 1 may be used as a base in a system of logarithms. However, only two bases are widely used in practice. The common, or Briggs, system of logarithms, named for Henry Briggs (1556-1631), employs the base 10 and is used for ordinary computations. The natural, or Napierian, system of logarithms, named for John Napier (1550-1617), is generally used in calculus and theo- retical work, and employs the more convenient irrational base e = 2.71828 In this book, when the base is not indicated, it is understood to be 10. Thus, log n means logi n, and the word logarithm will mean common logarithm unless otherwise stated. 5-4. COMMON LOGARITHMS In Table 5-1, we begin with a list of powers of 10, give equivalent logarithmic forms, and from these determine the form of the loga- rithm of a number that is not an exact power of 10. It should be mentioned that the logarithm is an increasing function; that is, as n increases, log n increases. Another way of stating the condi- tions is to say that if a> b then log a > log 6. 106 Logarithms Table 5-1 Sec. 5-4 Exponential form Logarithmic form Logarithm of the number 10 3 = 1000 log 1000 = 3.000 <-log354. = 2 + decimal 10 2 = 100 log 100 = 2.000 <-log 35.4 = 1 + decimal 10 1 = 10 log 10 = 1.000 <-log3.54 + decimal 10° = 1 logl 0.000 <-log 0.354 = — 1 + decimal io- 1 = 0.1 log 0.1 = - 1.000 <-Iog 0.0354 = — 2 + decimal io- 2 = 0.01 logO.Ol = - 2.000 <-log 0.00354 = — 3 + decimal io- 3 = 0.001 log 0.001 = - 3.000 From Table 5-1 it can be seen that the following statements are true: The logarithm of an integral power of 10 is an integer. The logarithm of a number which is not an integral power of 10 consists of two terms or parts : an integral part, called the charac- teristic; and a positive or zero decimal part, called the mantissa, which is determined from a table of mantissas. Thus, since log 10 = 1 and log 100 = 2, we may expect the loga- rithm of any number between 10 and 100, that is, a number between 10 1 and IO 2 , to be 1 plus a positive decimal part. For example, we shall find that the logarithm of 35.4, which number lies between 10 and 100, is equal to 1.5490, to four decimal places. In this case, the characteristic is 1 and the mantissa is .5490. 5-5. RULES FOR CHARACTERISTIC AND MANTISSA A study of Table 5-1 reveals that the characteristic changes as the position of the decimal point changes in the sequence of digits 0035400. The first entry in the column headed "Logarithm of the number" is log 354 = 2. + decimal. Sec. 5-5 Logarithms 1 07 In the number 354, or 354.0, the decimal point is two places to the right of the first non-zero digit, 3 (reading from left to right) ; the corresponding characteristic is 2. The second entry is log 35.4 = 1. + decimal. In this number, 35.4, the decimal point is one place to the right of the first non-zero digit (reading from left to right) ; the corre- sponding characteristic is 1. Similarly, we note that the zero characteristic corresponds to the position of the decimal point immediately following the first non- zero digit. This position of the decimal point is called the standard position. We may now formulate the following rule for characteristics : Rule for Characteristics. If the decimal point is in standard posi- tion, the characteristic is zero. For every other position of the decimal point, the characteristic is equal to the number of places the decimal point has been shifted from the standard position. The characteristic is positive if the shift is to the right, and is nega- tive if the shift is to the left. We shall now see that the mantissa remains the same for all numbers having the same sequence of digits. Let us again consider the sequence of digits 0035400. Any number containing this sequence can be written 3.54 • 10 n , where n is a positive or negative integer or zero and depends on the position of the decimal point. Suppose that we consider the form log 3.54 = 0.5490. Then the logarithm of any number containing this sequence is log (3.54 • 10 n ) = log 3.54 + log 10" = n + log 3.54 = n + 0.5490. Thus, a shift of the decimal place in the number affects only the characteristic n, and the mantissa remains the same for the same sequence of digits. EXERCISE 5-2 In each of the problems from 1 to 16, find the characteristic of the logarithm oh the given number. 1. 34.63. 2. f 3.463. 3. 34630. 4. 268.1. 5. 0.1340. 6. 2637. 7. 0.00346. 8. tan 42°8'. 7 821 9. sin 63°41'. 10. 0.000001. 11. 378364. 12. ]^^' 13. cot 81°13'. 14. sin 84°53'. 15. cos 61°43'. * 16. sec 24 b 8'. 1 08 Logarithms Sec. 5-5 In each of the problems from 17 to 24, place the decimal point in the sequence of digits 7314 corresponding to the given characteristic. 17. 3. 18. - 2. 19. 0. 20. 1. 21. 6. 22. - 5. 23. - 3. 24.-1. 5-6. HOW TO WRITE LOGARITHMS As stated in Section 5-4, the mantissa of a logarithm is always positive or zero, whereas the characteristic may be a positive or negative integer or zero. A positive characteristic or a zero char- acteristic can readily be combined with a given mantissa. For example, the logarithm of 354 is written 2.5490. But when the characteristic is negative, say — fc, where 1 ^ k ^ 10, it is more con- venient to write it in the form (10 — k) — 10. Let us consider the logarithm of 0.00354. The characteristic is —3, but the mantissa is regarded as positive. We could write log 0.00354 = — 3 + 0.5490. For convenience in computation, however, we write log 0.00354 in the form (10 - 3) + 0.5490 - 10 = 7.5490 - 10, or 17.5490 - 20, and so on. Note. It would be incorrect to write log 0.00354 = — 3.5490, for this notation means — 3 — 0.5490 and would imply that the mantissa is negative. To perform certain computations, it is convenient to write the logarithm 7.5490 - 10 in the form -2.4510, which equals — 2 — 0.4510. It is important to note that the decimal part of the number —2.4510 is not the mantissa of the logarithm of 0.00354, since it is not positive. 5-7. HOW TO USE A TABLE OF MANTISSAS The following examples will illustrate the procedure in finding the logarithm of a number with the aid of a table of mantissas. The student should work through each example, determining the char- acteristic from the position of the decimal point in the number and determining the mantissa by referring to Table III at the end of this book. Example 5-7. Find log 46.7. Solution: The characteristic is -f 1. To find the mantissa, locate 46 in the column in the table headed N, and then go to the right to the column headed 7. Here we find the mantissa .6693. So the complete result is log 46.7 = 1.6693. Interpolation. If the number consists of more than three digits, the mantissa is found from Table III by means of interpolation. Since the method of interpolation is the same as that described in Section 3-10 for the table of trigonometric functions, there will be no further discussion of it here. Sec. 5-7 Logarithms 109 Example 5-8. Find log 0.03426. Solution: The characteristic is — 2. The mantissa is found by interpolation, since the number 3426 has more than three digits. It lies — of the way between the mantissas of 3420 and 3430, as shown in the accompanying tabulation: Number Mantissa ■ f f 3420 .5340 1 1 6 * 10 1 3426 .5340 + x J ) 13 3430 .5353 Since the difference between the mantissas of the two numbers in the table is 13, we have a x=^ (13) = 7.8. This is rounded off to 8, and the amount to be added to 0.5340 is given by x = 8. Hence, the mantissa is .5348 and log 0.03426 = 8.5348 - 10. Finding Antilogarithms. The number which corresponds to a given logarithm is called the antilogarithrn. That is, if \ogn = x, then n is the antilogarithrn of x and is written antilog x. Example 5-9. Find n, if log n = 1.8710. Solution: Search through the body of Table III to locate the mantissa .8710. The corresponding number, from the columns headed N and 3, is 743. Since the characteristic is 1, n = 74.3. Example 5-10. Find antilog 7.5349-10. Solution: The mantissa .5349 is not in Table III but lies between .5340 and .5353. To these correspond, respectively, numbers whose digits are 3420 and 3430. We may indicate the work in tabular form as follows: Number Mantissa 10 3420 3420 + x .5340 .5349 13 3430 .5353 From this, we see that * x_ _ 9_ 10 ~ 13 # Therefore, x = 6.9, or 7 after rounding off. Hence, the sequence of digits in the desired number is 3427. Since the characteristic is - 3, the untilogarithm of 7.5349-10 is 0.003427. 110 Logarithms EXERCISE 5-3 Sec. 5-7 In each of the problems from 1 to 30, find the common logarithm of the given number. 3. 105. 4. 0.0843. 5. 0.00621. 8. 12.3. 9. 0.354. 10. 0.0781. 13. log 7.03. 14. log 95.5. 15. log 695. 18. 0.003821. 19. 0.7777. 20. 7,437. 23. 0.08788. 24. 15.46. 25. cos 16°13'. 26. sin 10°18'. 27. tan 41°33'. 28. sec 64°16'. 29. cos 82°14'. 30. cob 31°16'. In each of the problems from 31 to 50, find the antilogarithm of the given number. 1. 35. 2. 98. 6. 9.63. 7. 23,100. 11. 0.0663. 12. 1,630. 16. log 6.31. 17. 0.007001 21. 3.142. 22. 1.414. 31. 1.6665. 32. 4.4857. 33. 9.4183-10. 35. 2.7024. 36. 7.7388-10. 37. 9.4409-20. 39. 1.8401. 40. 3.9552-10. 41. 2.4658. 43. 9.7367. 44. 4.996O-10. 45. 8.7863-10. 47. 0.6584. 48. 3.0150. 49. 5.0300-10. Solve for x in each of the following equations: 51. 10- = 4. 52. 10* = 2.019. 54. 10 01 = x. 55. ^/10 = x. 57. v"l0* = z. 58. 10 1 - 314 = x. 60. 10-*' 2 = 0.0123. 61. 10 1 -* = 0.2346. 34. 0.0645. 38. 6.3404-10. 42. 1.9501. 46. 9.8821-20. 50. 0.1504. 53. 10 2 * = 7.132. 56. 10^« = x. 59. 10-* = 0.003146. 62. 10 2 *- 3 = 0.6735. 5-8. LOGARITHMIC COMPUTATION The fundamental laws of logarithms given in Section 5-2 are applied in the following examples to illustrate the application of logarithms to computation. Example 5-11. Find the product (0.0246) • (1360). Solution: Let x = (0.0246) • (1360). Then log x = log 0.0246 + log 1360. log 0.0246 = 8.3909 -10 log 1360 = 3.1335 logo; = 11.5244-10 = 1.5244. Hence, by interpolation, we have x = 33.45. Sec. 5-8 Logarithms 1 1 1 Example 5-12. Evaluate (0.506)- 1 ' 3 . Solution: Let x = (0.506) -»'» = (05 q 6)1/3 • Then log x = log 1 - log (0.506) 1/3 = log 1 - (1/3) log 0.506 = log 1 - (1/3) (29.7042-30) = log 1 - (9.9014-10). log 1 = 10.0000-1Q 1/3 log 0.506 = 9.9014-1Q log x = 0.0986. Therefore, x = 1.255 by interpolation. Alternate Solution: Let z = (0.506) - 1 ' 3 . Then log x = - (1/3) log 0.506 = - (1/3) (29.7042-30) = - (9.9014-10) = - ( - 0.0986) = 0.0986. Therefore, x = 1.255 by interpolation. n i r to ^ i x (0.352) (1.74)2 Example 5-13. Evaluate =~^ — • ^0.00526 Solution: Let x denote the desired value. Then log x = log 0.352 + 2 log 1.74 - (1/3) log 0.00526. We find that log 0.352 = 9.5465-10, log 1.74 = 0.2405, and log 0.00526 = 7.7210-10. log 0.352 = 9.5465-10 (1/3) log 0.00526 = (1/3) (27.7210-30) 2 log 1.74 = 0.4810 = 9.2403-10. log numerator = 10.0275-10 log numerator = 10.0275-10 log denominator = 9.2403-10 log x = 0.7872. Interpolating, we have x = 6.126. 253 y 14 (253' 174 Solution: Let x = ( j=j j . Then log x = 1.14 [log 253 - log 174]. log 253 = 2.4031 log 174 = 2.2405 Then °' 1626 - log x = 1.14 (0.1626) = 0.1854. Therefore, x = 1.532. 112 Logarithms Sec. 5-8 EXERCISE 5-4 In each of the problems from 1 to 30, perform the indicated computation using logarithms. 1. (3.142)(2.718). 29.34 3, 683.5 5. ^(5.678) 2 . 7. \/(0.003468)\ 9. a V (5,321, OOP)* V (36,250) 4 11. (63.84)2(0.0134). 13. V(168.3) (14.21). 15. V(23,310) 2 -(20,180)2. (Hint: Factor the radicand.) t » /123.4V* 19, 21 1 • 3 • 5 • 7 « 19 • 31 2* 4- 8- 16 •32*64* V (8.013) (0.034) 23. (18,120) > (1,0 ^° 4 "" 1 > 25. </68l3 - (0.8123)- 3 ' 4 . 27. ^log 0.08614. 29. l(3.864)-3-i3 + (0.841)-o-wp'«. 2. (13.25) (26.80). 4. (0.8134) 1/3 . 6. (16.83) 3 '*. 42,860 ' 10. (4.313)(3,068)(0.000642). 12. (8.364)(321.5)+(- 42.63). 14. V(213.6) 2 (43.98)2. 1A /68.34y' 5 Mara?; • 18. 20, 3,6 42 y/ (21.36) 3 (1,083) 4 (0.0813) 3 1.63 V 0. 483 38, 22. (1.08)1°. 8103 24. 3,648 (1.03) 36 . 26. 0.083 ' 412 . 28. log 16.84 - ^483.6. 30. (83.14)" 4 - 3 - (0.8134)2/3 (0.6841)1/2 31. If e = 2.718, find log e, log \/e, log - , e* f and t. 32. Find the geometric mean of 564.3, 8634, 0.1349, 8.316, and 42.61. (Hint: See Problem 43, Exercise 5-1.) 33. Find the area of a circle of radius 6,381 feet. 4 34. The volume of a sphere is V = « rr z . Find the volume of a sphere of radius 3621 feet. 6 35. Find the radius of a sphere whose volume is 8423 cubic feet. 36. Find the length of a pendulum which makes one oscillation in 1 second, if g = 980 centimeters/sec 2 . (Hint: See Problem 41, Exercise 5-1.) Sec. 5-9 Logarithms 1 1 3 37. Find the area of a triangle with sides 6,384 feet, 5,680 feet, and 2,164 feet long. (Hint: See Problem 42, Exercise 5-1.) 38. The stretch s of a wire of length I and radius r by a weight m is given by the relationship s = — %r > where g is the gravitational constant and k (Young's modulus) is a constant for a given material. Find how much a copper wire of length 120 centimeters and of radius 0.040 centimeters will be stretched by a weight of 6,346 grams, if g = 980 and k is 1.2 • 10 12 for copper wire. 39. The current i flowing in a series circuit with a resistance of R ohms and L henrys t seconds after the source of electromotive force is short-circuited is given by the relationship i = Ie~ RtlL , where I is the current flowing in the circuit before the short circuit. If i = 10 amperes, R = 0.1 ohms, t = 0.25 seconds, and L = 0.05 henrys, find /. (Take e = 2.718.) 40. If n is a positive integer, n! has been defined as the product 1 • 2 n. When n is very large, it is difficult to compute this product. However, Stirling's formula gives (approximately) n! = n n e~ n -\/2irn. Use this formula to estimate 9!, and compare the result with the true value which you should calculate exactly. Do the same with 30!. (Hint: log (n!) = n log n — n log e + 2 log 2 + ~ log 7T + - log n). 41. If the rate of depreciation r per year is constant, the scrap value S after n years of a machine with first cost C is given by the formula S = C(l — r) ft . Find the scrap value after 10 years of a machine which originally cost $10,000, if 20 per cent per year is written off as depreciation. 5-9. CHANGE OF BASE It is sometimes desirable to change from one logarithmic base to another. Suppose there is available a table of logarithms to some known base b (say 10, for exampte), and we wish to find the loga- rithm of a number n to some other base a. We then let x = log* n ; whence, by definition, n = 6*. Similarly, if we let y = log n, then we have n = a y . It follows that a y = b*, and our problem reduces to solving this equation for y. Taking the logarithm of both sides to base 6, we have t y log* a = x log& b. But log6 6 = 1. Therefore, (5-5, "»-Eji- 1 1 4 Logarithms Sec. 5-9 Example 5-15. Find log* 125, where e = 2.7183, by using a table to the base 10. . Solution: By (5-5), loge 125 = -p 6 logio e It is usually easier to multiply than to divide. Since division by logio e is a fairly frequent operation in practical work, it should be noted that ^aoao = 2.3026, and the result can be obtained by multiplying by 2.3026 instead of dividing by 0.4343. Thus, log. 125 = (2.0969) (2.3026) = 4.828. EXERCISE 5-5 Find each of the following logarithms by using a table of common logarithms: L log« 10. 2. log 9 100. 3. log2 e. 4. log e tt. 5. log* e. 6. log* 10. 7. log 2 64. 8. log 20 1000. 9. log 20 100. 10. logioo 64. 11. log e 8. 12. logo.i 50. 13. logi25 1000. 14. loge 20. 15. logo.02 0.04. 16. logiooo 100. 6 Right Triangles and Vectors 6-1. ROUNDING OFF NUMBERS Numbers that arise in the applications of trigonometry are usually not exact, but are sufficiently accurate for a given purpose. Numbers of this kind are called approximate numbers, and the degree of accuracy of such a number is indicated by how many significant figures it contains. Reading from left to right, the significant figures in a number are the digits starting with the first non-zero digit and ending with the last non-zero digit, unless it is definitely specified that the zeros on the right are significant. Thus, in the numbers 2.405, 0.002405, and 240500, the digits 2, 4, 0, and 5 are significant figures. The zeros after the 5 in 240500 may or may not be significant figures. When it is desired to indicate whether final zeros are significant or not, scientific notation is often used. Thus, in 2.405 • 10 5 , the last significant figure is 5 ; in 2.40500 • 10 5 , the final two zeros are regarded as significant. To round off a number in which the last desired significant figure is in the units place or in any decimal place, drop all digits that lie to the right of the last significant figure. It is sometimes necessary also to increase the last digit in the retained part by 1. If the first digit in the dropped part is less than 5, the last digit in the retained part is left unchanged. If the first digit in the dropped part is greater than 5 or if that digit is 5 and it is followed by digits other than 0, the last digit in the retained part is increased by 1. Whten the dropped part consists of the digit 5 alone or the digit 5 followed only by one or more zeros, we shall use the following procedure as an arbitrary rule in this book: If the last digit retained is odd, this digit is increased by 1; if it is even, it is left unchanged. This rule, although popular, is inferior 115 1 1 6 Right Triangles and Vectors Sec. 6—1 to common-sense rules in many cases. For example, if .245, .165, .485, and .725 are to be rounded off to two decimal places and then added, it would be more sensible to round off two of the numbers in one direction and two in the other direction. To round off a number in which the last significant figure will lie to the left of the units place, first drop all digits to the right of the place occupied by the last significant figure, and replace each dropped digit to the left of the decimal point by a zero. Also, either leave the last digit of the retained part unchanged or increase that digit by one, in accordance with the directions just given for dropping only a decimal part. For example, if 2533.62 is to be rounded off to three significant figures, the result is 2530 ; and if 487,569 is to be rounded off to three significant figures, the result is 488,000. There are two rules that are generally adopted by computers in working with approximate numbers in order to guard against retaining figures that may indicate a false degree of accuracy : 1. In adding or subtracting approximate numbers, round off the answer in the first place at the right in which any one of the given numbers ends. 2. In multiplying or dividing approximate numbers, round off the answer to the fewest significant figures found in any of the given numbers. The numbers entering a problem involv- ing multiplication or division may be rounded off before the computation is begun. If these numbers are rounded off, they should have one more significant figure than the answer is to have. While these rules point in the right direction, it should be men- tioned that rounding off computed quantities to as many significant figures as there are in the given numbers does not necessarily produce the degree of accuracy implied by the results. The subject of accuracy of computation with approximate numbers is somewhat complicated and beyond the scope of a book at this level. 6-2. TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES One of the simplest, yet important, applications of trigonometry is in the solution of right triangles. A right triangle has, in addi- tion to the 90° angle, five other parts. These are two acute angles and three sides. If we know the length of any two sides, or either acute angle and any one side, the triangle can be solved; that is, the unknown parts can be found. Sec. 6-3 Right Triangles and Vectors 117 hV •fl c s a A b C " To solve problems involving parts of tri- angles, we shall find it helpful to be able to express the trigonometric functions of an acute angle A of a right triangle ABC in terms of the sides of that right triangle. To derive suitable relationships, let us place the acute angle in standard position, as shown in Fig. 6-1. In the right triangle ABC the side AC, which is the abscissa of the point B, becomes the side adjacent to the angle A ; the side CB, which is the ordinate of B, becomes the side opposite to angle A ; and the side AB, which is the radius vector to B, becomes the hypotenuse. If we let the lengths of the side adjacent, the side opposite, and the hypotenuse be represented by the symbols b, a, and c, respectively, we may express the six functions of the acute angle A in terms of a, b, and c as follows : . A __ side opposite __ a Sin Jx — "-; ~ — hypotenuse c — ^de adjacent _ 6 hypotenuse "~ c side opposite __ a (6-1) (0-2) cos A (6-3) (6-4) (6-5; (6-6) tan A = esc A sec A cot A = side adjacent hypotenuse side opposite hypotenuse side adjacent side adjacent side opposite a By using (6-1) to (6-6), we can express the trigonometric func- tions of an angle of a right triangle without reference to any coordinate system, since the ratios of the sides remain the same regardless of the position of the triangle. 6-3. PROCEDURES FOR SOLVING RIGHT TRIANGLES When solving a right triangle in which two parts are known, it is advisable to arrange the work systematically and to follow £ definite procedure consisting of the following steps : 1. Draw a figure reasonably close to scale, and indicate the known parts. 2. Write an expression containing a trigonometric function which involves the two known parts and one unknown part. 1 1 8 Right Triangles and Vectors Sec. 6—3 3. Find the selected unknown part from this equation. 4. Find all other unknown parts of the triangle by a similar procedure. 5. Check all results. Whenever possible, select a trigonometric function that gives a solution by means of a multiplication rather than a division. In the following illustrative examples, the acute angles are represented by the letters A and B, and the right angle is ^^ a -658 denoted by C, while the small letters a, b, and c, respectively, represent the sides opposite them. Example 6-1. Solve the triangle ABC, if A = 28°20' and a = 658. Solution: The triangle is drawn approximately to scale in Fig. 6-2. The unknown parts are the angle B and the sides b and c. Since A + B = 90°, we have B = 90° - 28°20' = 61°40'. To find the side b, we may apply either (6-3) or (6-6), since both equations involve the unknown b and the known parts A and a. We shall use cot A = - because it enables us to proceed to the solution by means of a multiplication rather than a division. Since A = 28°20' and a = 658, we have 4 = cot28 ° 20 ' Then b = 658 cot 28°20 / = (658) (1.855) = 1220.59. Xn this example, we take b equal to 1221. This result is rounded off to four digits. To find the side c, we shall use (6-4), or esc A = - • We have, therefore, esc 28°20' = 7^3 • 658 Hence, c = 658 esc 28°20 / = (658) (2.107) = 1386. 1221 To check, we may use the relation cos A = -r^^ — 0.8802. Hence, A = 28°20'. Checking by means of the Pythagorean theorem yields the result b 2 = c 2 - a 2 = (c - a) (c+a) = (728) (2044) = 1488232, whereas b 2 = (1221) 2 = 1490841. These values of b 2 agree when they are rounded off to three significant figures. Note. In most situations where we must solve triangles, we are dealing with measured quantities, which are necessarily approxi- mate. Therefore, our answers can be no more accurate than the Sec. 6-3 Right Triangles and Vectors 1 19 data we begin with. If the original data are approximate, our answers must be rounded off to the degree of accuracy indicated by the data. In example 6-1, for instance, the answers may be given as b = 1220 and c = 1390, both rounded off to three significant figures. Fig. 6-3. Example 6-2. Solve the triangle ABC, if b = 250 and c = 371. Solution: The conditions are shown in Fig. 6-3. Since - = cos A, we have cos A 2^0 c = =£ = 0.6738. Therefore, A = 47°38' and B = 90° - 47°38' = 42°22'. To find a, we have a choice of combining the unknown a with either b or c; hence, we may use either -r = tan A or - = sin A . We shall illustrate, in order, the computation with each of these equations, thus providing a check on our work. Using (6-3), we have ^o = tan47 ° 38 - Hence, a = 250 tan 47°38' = (250) (1.096) = 274.0, or 274 when rounded off to three figures. Using (6-1), we have ~L = sin 47 o 38 / b Hence, a = 371 sin 47°38' = (371) (0.7388) = 274.09 or 274 when rounded off to three figures. EXERCISE 6-1 In each of the problems from 1 to 16, solve the right triangle. 1. a = 12, A = 33°. 2. b = 168, A = 38°16'. 3. b = 62.4, B = 71°10'. 4. a = 42, c = 76. 5. a = 3.187, b = 6.249. 6. 6 = 63.21, B = 83°36\ 7. a = 4.318, B = 67°16'. 8. b = 827.6, c = 963.4. 9. a = 9.863, A = 36 21'. 10. 6 = 16.32, 5 = 87°10\ 11. 6 = 78.21, A = 43°17'. 12. a = 43.21, c = 63.75. 13. a = 123.6, 6 = 783.1. 14. a = 36.83, A = 57°44'. 15. 6 = 2.312, B = 40°57'. 16. a = 389.3, 6 = 62.34. ? 17. A wire stretches frorrj a point on level ground to the top of a vertical pole. It touches the ground at a point 15 feet from the foot of the pole and makes an angle of 63° with the horizontal. Find the height of the pole and the length of the wire. 18. A ladder 40 feet long rests against a vertical wall. If its footjs 5 feet from the base of the wall, what angle does it make with the ground? 120 Right Triangles and Vectors Sec. 6-3 19. A ladder 65 feet long is placed so that it will reach a window 35 feet above the ground on one side of a street. If the foot of the ladder is held in the same position and the top is moved to the other side of the street, it will reach a window 28 feet above the ground. How wide is the street from building to building? •20. The grade of a hill is the tangent of the angle the hill makes with the horizontal. Find the grade of a hill which is 275 feet long and which rises 120 feet. 21. To find the width of a river, a surveyor sights on a line across the river between two points A and B on opposite banks of the river. He then runs a line AC perpendicular to AB. He finds that AC is 250 feet and angle ACB is 42°17'. How wide is the river? 22. Find the length of a side of a regular hexagon and the radius of the inscribed circle, if the radius of the circumscribed circle is 10 feet. 23. An airplane rises 560 feet while flying upward for 2,387 feet along an inclined straight-line path. What is the angle of climb? 24. A pendulum 4.5 inches in length swings through an arc of 28°. How high does the bob rise above its lowest position? 25. A man 6 feet tall is walking along a straight horizontal path directly away from a lamp post 10.5 feet high. How far is he from the post at a certain instant when his shadow is 5 feet long? Horizontal Fig. 6-4. 6-4. ANGLES OF ELEVATION AND DEPRESSION Let the line AB in Fig. 6-4 be a level or horizontal line, and let an observer at the point A see an object at the point C. If the object C is above the horizontal line AB, then the angle BAC meas- ured up from the horizontal to the line of sight AC is called the angle of elevation to C from A. If the object C is below the hori- zontal line AB, then the angle BAC measured down from the* horizontal to the line of sight AC is called the angle of depression to C from A. Example 6-3. From a point on the ground 300 feet from the base of a building, the angle of elevation to its top is 22° 10'. How high is the building? Solution: In Fig. 6-5, we have b = 300 and A = 22°10'. By (6-3), ~g = tan 22°10'. Hence, a = 300 tan 22°10' = (300) (0.4074) = 122.22. So the build- ing is 122 feet high. Sec. 6-5 Right Triangles and Vectors **E 121 *»e Fig. 6-5, 6-5. BEARING IN NAVIGATION AND SURVEYING In marine and air navigation and in surveying, the direction in which an object is seen is expressed by the bearing or azimuth of the line of sight from the observer. The bearing of a line is the acute angle which its direction makes with a meridian or north- south line. Such angles are sometimes called quadrant angles, or quadrant bearings. To describe the bearing of a given direction, we first write the letter N or S, then the acute angle, and finally the letter E or W. The letters depend on the quadrant in which the given direction falls. Thus, the bearings of the lines OA, OB, and OC in Fig. 6-6 are N 60° E, S 37° W, and N 45° W, respectively. The azimuth of a line differs from its bearing only in that the azimuth is the angle measured from 0° at north in a clockwise direction. An azimuth may have any value between 0° and 360°. Thus, in Fig. 6-6 the azimuths of the lines OA, OB, and OC are 60°, 217°, and 315°, respectively, measured clockwise from the north. This method of measuring directions is coming into more frequent use than that of quadrant bearings. We note also that the term bearing is often used instead of azimuth. Thus, we may speak of the bearing of an object regardless of whether we mean azimuth or bearing as here defined. Example 6-4. A ship heads due east from a dock at a speed of 18 miles per hour. After traveling 30 miles it turns due south and continues at the same speed. Find its distance and bearing from the dock after 4 hours. * Solution: In Fig. 6-7, let A be the point at which the dock is located, let B be the point where the ship turns south, and let C be the position of the ship after 4 hours. 30 5 Since the number of hours required to travel from A to B is ^ = - > the number 5 7 7 of hours spent in travel from B to C is 4 - = = ^ • Hence, a =" 18 • ^ = 42. 122 Right Triangles and Vectc From the figure, 42 tan = r^ and sec = Therefore, = 54°28', and Sec. 6-5 30 b = 30 sec 54°28' = (30) (1.721) = 51.6. Hence, the distance from the dock is 52 miles and the bearing of the line AC is (44°28' or S 35°32' E. 6-6. PROJECTIONS Often it is desirable to consider direction along a line segment. Thus, if Pi and P 2 are the end points of a segment, we shall under- stand PiP 2 to mean the directed segment from P x to P 2 , the direc- tion being specified by the order in which the end points are named. The non-negative length of the segment PiP 2 is denoted by |PiP 2 |. Frequently a directed segment PiP 2 may lie on a line, such as a coordinate axis, on which a positive direction has been specified. Then the positive direction on the line may agree with the direction from Pi to P 2 , or the two directions may be opposite to each other. The directed length of the segment PiP 2 is equal to |P X P 2 | when the directions agree or when Pi and P 2 coincide and is equal to -|PiP 2 | when they disagree. Since the context will make the meaning clear, we shall designate the directed length of the segment PiP 2 also by PiP 2 . IF mr 2 ) C(o,y x ) J*2 ( x z*yz) / A(x v Q) B(x 2> 0) Fig. 6-9. *»x We recall that the projection of a point on a given line is the foot of the perpendicular dropped from the point to the line. If A in Fig. 6-8 is the projection of Pi on the line I, and if B is the projec- tion of P 2 on I, then the directed segment from A to B is the projection on I of the directed segment P\P 2 . We draw P X M parallel to I, or perpendicular to P 2 B, to show the angle between I and PA. Sec. 6-6 Right Triangles and Vectars 123 We shall now assume that a positive direction has been specified on the line I. Then, since PiM = AB, it follows immediately from trigonometry that AB = | PiP 2 I cos 0, where 6 is the acute angle between the positive end of I and the positive half -line determined by the directed segment P1-P2. In Fig. 6-8 it is considered that I is positively directed toward the right. The result just given can be applied, as seen in Fig. 6-9, in find- ing the projections of P\P 2 upon the coordinate axes. The directed lengths of the projections upon the cc-axis and the 2/-axis are, respectively, (6-7) AB = I P1P2 I cos 0, (6-8) CD = I P1F2 I sin 0, where is the angle between OX and PiP 2 , as shown in Fig. 6-9. When the coordinates of the end points of the segment P\P 2 are known, the projections AB and CD are readily expressed in terms of these coordinates. From the definitions of horizontal and vertical distances given in Section 2-2, it follows that (6-9) AB = #2 — #i and CD = 2/2 — j/i. i iV C (0,2) ■ i P x (3,2) #(7,0) U(3,0)'\ IX0.-5) ■ P 2 (7,- Fig. 6-10. Fig. 6-11. Example 6-5. What kre the projections of the segment P1P2 on the axes, if Pi = (3, 2) and P 2 = (7, - 5)? Solution: In Fig. 6-10, AB = z 2 - 21 = 7 - 3 = 4, and CD = t/ 2 - 2/1 = -5-2=-7. Since AB is -f 4, we know that AB is directed to the right. Also, since CD = - 7, we know that CD is directed downward. 1 24 Right Triangles and Vectors Sec. 6-6 Example 6-6. A ladder 12 feet long leans against the side of a house and makes an angle of 67° with ground. Find its projections on the ground and on the side of the house. Solution: The conditions are represented in Fig. 6-11. Let I = 12 be the length of the ladder. The required projections are found as follows. The projection on the ground is given by x = I cos 6 = 12 cos 67° = 12 (0.3907) = 4.7. The projection on the side of the house is given by y = I sin = 12 sin 67° = 12 (0.9205) = 11.0. EXERCISE 6-2 1. Two points A and B are 5,000 feet apart and at the same elevation. An airplane is 10,000 feet directly above point A. Find the angle of depression from a horizontal line through the airplane to point B and the airplane's distance from point B. 2. The Washington monument is approximately 555 feet high. Find the angle of elevation to the top of the monument from a point that is 621 feet from the base of the monument and at the same elevation as the base. 3. If a kite is 130 feet above the ground and 150 feet of string is out, find the angle of elevation to the kite, assuming the string to lie on a straight line, 4. Find the angle of elevation to the sun if a flagpole 95 feet high casts a shadow 63 feet long on horizontal ground. 5. A boat leaves its dock and heads N 52° W for 4 hours at 14 knots (1 knot = 1 nautical mile per hour = 6,080.4 feet per hour). It then turns and heads N 38° E for 3 hours at 16 knots. Find the boat's final bearing and distance from the dock. 6. The grade of a certain railroad bod is 0.1095. How many feet does a locomotive rise while traveling 175 feet along the track? 7. An approach must be built up to the end of a bridge which is 40 feet above ground. If the approach is to have a 10% grade and the original ground is assumed to be level, how far from the end of the bridge must the approach start? 8. A surveyor wishes to find the distance between two points A and B separated by a lake. He finds a point C on the shore of the lake such that angle ACB is 90°. He measures AC and BC and finds that AC is 640 feet and BC is 285 feet. How far apart are A and B? 9. A smokestack is 175 feet from a building. From a window of the building the angle of elevation to the top of the stack is 28°10'. The angle of depression to its base from the same window is 24°30'. Assuming that the ground is level, find a) the height of the window above the ground and b) the height of the smokestack. 10. Two ships leave the same port at the same time. One travels N 42° E at 25 knots. The other travels S 48° E at 33 knots. How far apart are the two ships after 4 hours? Sec. 6-7 Right Triangles and Vectors 125 6-7. SCALAR AND VECTOR QUANTITIES We shall at this point find it necessary to distinguish carefully between two kinds of quantities, namely scalar quantities and vector quantities. A scalar quantity is a quantity whose measure can be fully described by a number. It is a quantity which can be measured on a real number scale. For example, temperature is a scalar quan- tity, measured on the scale of a thermometer. Also we shall define the scalar components of the segment PiP 2 to be the projections on the coordinate axes, or the directed lengths x 2 — Xi and y 2 — Vi given by (6-9) in Section 6-6. The student should note, however, that the scalar components of P 2 P\ are not equal to those of P1P2. The components of P 2 P\ are x x — x 2 and y x — y 2 and are the nega- tives of the respective components of PiP 2 . A vector quantity, or simply a vector, is a quantity possessing both magnitude and direction. A vector may be represented by any one of a set of equal and parallel line segments. Algebraically, a vector is fully described by the scalar components of any segment representing it ; all such segments have the same components. We shall, in fact, call these the scalar components of the vector and shall £\ — i — i n \S. . . . IY i >*i — i i i i i O. Fig. 6-12. «D IF cr • i i i i i i i i i i i i — >-x -*x Fig. 6-13. enclose them in brackets. In Fig. 6-12, three representations of the vector [4, 3] are shown. The arrowhead indicates the order in which the end points of the line segment are named. We may denote a vector by a single letter in boldface type, say v, or may represent it geometrically by any one of the segments, such &s AB y OP, or CD. Here A, O, and C represent the initial points of the three segments, while the terminal points at the arrowheads are B, P, and D. Actually, any other segment with the same magnitude and direction could have been selected to represent the vector v. Instead of using a single letter in boldface type to denote a vector represented by a segment P\P 2 , we may also use the notation PiP* 1 26 Right Triangles and Vectors Sec. 6-7 Properties of Vectors. We have seen that a vector is unambigu- ously represented by any one of a set of equal and parallel line segments. Hence, any point may be taken as the initial point of a segment which represents a vector. If the origin is so chosen, the coordinates of the end^point P(x,y) are actually the scalar components of the vector OP. We can therefore give the following simple definition of the magnitude of a vector and its expression in terms of scalar components : Definition. The magnitude, or length, of a vector v is the length of any one of the segments representing v. Let v = [>i, v 2 "\ be the vector represented by OP in Fig. 6-13, where v x and v 2 are scalar components. Then we have |v| = \OP\. Since OP is the hypotenuse of a right triangle, (6-10) | v | = VV + *>2 2 . In case the scalar components are given by the coordinates (x, y) of the end point P of the segment, the magnitude of the vector is y/x*+y 2 . From the definition of a vector, it follows that two vectors are equal if and only if their respective scalar components are equal. For example, u = [u lt u 2 ] equals v = [v lt v 2 ] if and only if u x = v x and u 2 = v 2 . We also define a special vector = [0, 0] to be the zero vector. It corresponds to the exceptional case in which P 2 coincides with Pi and may be considered as represented geometrically by a seg- ment of length zero, that is, by a point. The zero vector may be regarded as having any direction whatsoever. If a vector v. = \v l9 v 2 "] is given, the vector — v = [— v u —^2] is defined to be the negative of v. Thus, if P X P 2 = v denotes a vector represented by the segment PiP 2 , then P 2 P X = -v denotes a vector having the same length as P\P 2 but oppositely directed, namely, from P 2 to Pi. We note that - (- y) = v. A unit vector is defined as a vector whose magnitude is unity. If u = \u l9 ^2] is any non-zero vector, then j-5- = L^L. , -^-1 is a unit vector. Multiplication of a vector by a scalar is performed by multiplying the magnitude of the vector by the absolute value of the scalar, maintaining the direction of the vector if the scalar is non-negative and reversing it otherwise. Thus, by k[u u u^\ we mean the vector Sec. 6-7 Right Triangles and Vectors 1 27 [ku lt ku 2 ]. Using the equation [m, U2] = k |~ > ^1 1 we can express any vector v = [i^, w 2 ] as proportional to a unit vector v/k if we choose k = V^i 2 + ^2 2 . Changing v to a unit vector v/fc is called normalizing v. Sums and Differences of Vectors. Of the many questions which arise in the study of vectors, the one of greatest importance for us at present concerns the addition of vectors. The sum, or resultant, of two vectors is defined to be the vector which has for its scalar components the sums of the scalar components of the two vectors. Thus, the sum of the vectors u and v is given by the relationship (6-11) U + V = [Wi + Vi, U 2 + V2I It is important to note that some of the laws of the algebra of numbers also hold for vectors. Thus, the commutative law is u + v = v + u. Also, the associative law of addition is (u + v) + w = u + (v + w). And, for every vector v, v + ( - v) = 0. That these laws are satisfied for vectors can be shown geometrically or can be seen from (6-11) by virtue of the known laws of addition of real numbers. In terms of components, the rule for multiplication by a scalar is given by fcv = [kviy kv2]. In consequence of this definition, the following algebraic laws are satisfied : k(vi + y) = fcu + k\; (k + m)u = fcu + ma) k(mu) = (fcm)u. To find the sum of two vectors geometrically, we proceed as indicated in Fig. 6-14. This graphical representation of the sum of two vectors by means of the triangle construction was probably suggested by the behavior of physical quantities represented by vectors, such as forces, displacements, velocities, and accelerations. Their addition is effected by the triangle law or the parallelogram law. 128 Right Triangles and Vectors Sec. 6-7 *»X Fig. 6-14. Fig. 6-15. The following steps indicate the actual procedure. 1. Select a segment representing one of the vectors, say u, with its initial point at the origin of the coordinate system. 2. Place the initial point of a segment representing the second vector, say v, at the terminal point of the first segment. 3. Draw the segment from the origin to the terminal point of the segment for v. This segment represents the resultant vector, u + v. The difference u - v of two vectors is defined in a manner analogous to that used in defining the difference of numbers. Thus, u — V = u + (— v). If u = [u u v^\ and v = [v u v 2 ~\ , then U — V = [Ui — Vi, U2 — V2]. To find the difference u — v of two vectors geometrically, we proceed as indicated in Fig, 6-15. The steps are as follows : 1. Place segments representing u and v with their initial points at the origin. 2. Place the initial point of a segment representing — v at the terminal point of the segment representing u. Note that the segment representing — v is parallel and equal in length to that for v, but has the opposite direction. 3. Draw the segment from the origin to the terminal point of the segment for —v. This segment represents the vector u — v. Example 6-7. Find the sum and difference of the vectors u = [5, - 3] and ▼ = [-2,1]. Solution: The required vectors are n + ▼ = [5 - 2, - 3 + 1] = [3, - 2], and u-v = |5-(-2), -3-l] = [7, -4]. Sec. 6-*7 kight Triangles and Vectors 129 Example 6-8. Express u = [4, 3] as proportional to a unit vector. Solution: The expression representing the vector is U = [4,3]=5[|,|]. [4 3"! - t - . To check, T4 31. note that the magnitude of U > r is To add vectors analytically, we first find the scalar components of the vectors to be added. From Section 6-7, we know that if segments represent- ing u and v make angles a and (3, respectively, with the #-axis, the scalar com- ponents are given by the projections upon the coordi- - nate axes. Thus, as shown in Fig. 6-16, u x = | u | cos a, and v x = | v | cos j3 7 fac,i>y,), Fig. 6-16. Uy = I U | Sin OL, vy = | v | sin j8. Then the components of the resultant will be given by the algebraic sums X = u z + v x , Y = Uv + vu. Thus, the magnitude of the resultant R is given by | R | = V^ 2 + Y 2 . Also, the direction angle 6 satisfies the relationship Y tan 6 = -y • To determine the quadrant of correctly, it is important to keep in mind and use the correct signs of X and Y. For example, tan = -^-i = 1 might lead one to an incorrect value 45° for $ instead of the correct third-quadrant angle 225°. The student should note that this (analytic) procedure and the geometric procedure (parallelogram law) give the same results. This can be shown by appropriately combining the scalar components of the vectors u, v, u + v, and u - v in Figs. 6-14 and 6-15, where we illustrated the geometric procedure. 130 Right Triangles and Vectors Sec. 6-7 Fig. 6-17. *~E Example 6-9. A block weighing 500 pounds rests on a smooth plane making an angle of 25° with the horizontal, as indicated in Fig. 6-17. What force, parallel to the plane, is necessary to hold the block in position? Solution: Let the block be at A. The force due to the weight mayjbe represented by a vector acting vertically downward through A, such as vector AC. The com- ponent of AC which is parallel to the inclined plane and which must be overcome is represented by the projection AB of AC on the inclined plane. Since angle BAC = 90° - 25° = 65°, AB = 500 cos 65° = (500) (0.4226) = 211.3. Hence, a force of 211 pounds must be applied parallel to the inclined plane to keep the block from sliding. (It is assumed that three significant figures are appropriate.) Example 6-10. Find the magnitude and direction of the resultant of a force of 110 pounds acting in the direction S 46°27' E and a force of 100 pounds acting in the direction N 29°14' W. Solution: The given forces are represented by vectors in Fig. 6-18. Let F denote the magnitude of the resultant force and 6 the angle it makes with OE. Since 90° - 46°27 / = 43°33' and 90° - 29°14' = 60°46', the components F x and F y of the resultant are found as follows: F. = 110 cos 43°33' - 100 cos 60°46' = (110) (0.7248) - (100) (0.4884) = 79.73 -48.84=30.89; F v = 110 sin 43°33' + 100 sin 60°46' = - (110) (0.6890) + (100) (0.8726) = - 75.79 + 87.26 = 11.47. Now tan 6 F x 11.47 = 0.3713, and esc 6 = F 30.89 ~ "•"• *"' m ~ wv v ~" 11.47 Therefore = 20°22', and F = 11.47 esc 20°22' = (11.47) (2.873) = 32.95. Hence, the magnitude of F is 33 pounds, and its direction is N 69°38 / E. Sec. 6-8 Right Triangles and Vectors 131 EXERCISE 6-3 L Draw a diagram showing three different segments representing each of the given vectors. Find the magnitude of each vector. a. [3, 4]. b. [12, 5], c. [-2, 4]. d. [- 3, - 5]. e. [1, - 1]. f. [5, 3]. g. [0, 4]. h. [0, - 1]. 2. Express each of the vectors in Problem 1 in terms of a unit vector. 3. In each of the following cases, add the given vectors. Find the magnitude and direction of the resultant. a. [1, 1] and [2, 3]. b. [4, - 2] and [1, - 10]. c. [2, 0] and [ - 6, 3]. d. [1, 2], [4, 3], and [0, 7]. 4. In Problem 3, parts (a), (b), and (c), subtract the first vector from the second, and find the magnitude and direction of the resultant. 5. In Problem 3(d), subtract twice the third vector from the sum of four times the first vector and twice the second vector, and find the magnitude and direction of the resultant. 6. A force of 40 pounds acts at an angle of 63° with the horizontal. What are the vertical and horizontal components of the force? 7. If a ship sails N 48° W at 30 knots, what are its westward and northward components? 8. One force of 28 pounds acts vertically upward on a particle. Another force of 43 pounds acts horizontally on the particle. What is the magnitude of the resultant force, and what is its direction? 9. A barrel weighing 160 pounds rests on a smooth plane which makes an angle of 22° with the horizontal. Find the force parallel to the plane necessary to keep the barrel from rolling down the plane. 10. Three forces act on a particle. One of 50 pounds makes an angle of 25° with the horizontal; a second of 60 pounds makes an angle of 50° with the horizontal; and the third of 75 pounds makes an angle of 230° with the horizontal. Find the magnitude and direction of the resultant force. 11. Four forces act on a body. The forces are 30, 45, 50, and 65 pounds, and they make angles with the horizontal of 25°, 160°, 240°, and 330°, respectively. Assuming that all the forces lie in the same vertical plane, find the magnitude and direction of the force necessary to hold the body in equilibrium. The required force is equal in magnitude and opposite to the resultant of the given forces. 12. A shell is fired at an angle of elevation of 37°. Its initial velocity is 2,500 feet per second. Find the horizontal and vertical components of its initial velocity. 6-8. LOGARITHMS OP TRIGONOMETRIC FUNCTIONS So far in this chapter, we have considered the use of a table Qf natural trigonometric functions and have solved various problems involving right triangles. In many problems, however, the compu- tation is greatly facilitated by the use of logarithms to perform the numerical operations. For this purpose the values of the logarithms of the trigonometric functions are required. Table III might be used to obtain the logarithms of the functions found in Table II, 132 Right Triangles and Vectors Sec. 6-8 but the work is considerably lessened by the use of Table IV at the end of this book, which gives the logarithms of the trigonometric functions at once. Table IV is a four-place table giving the logarithms of functions at intervals of 10 minutes from 0° to 90°. For the sine and cosine of any angle between 0° and 90°, the tangent of any angle between 0° and 45°, and the cotangent of any angle between 45° and 90°, the value of the function is less than 1; hence, the logarithms of these functions are negative, and we must write —10 after the tabulated entry. For the sake of uniformity, 10 has been added to each of the other entries in the table. In using the table, therefore, 10 must be subtracted from every entry. The method of using Table IV is similar to that described for Table II and will be illustrated by the following examples. Example 6-11. Find log sin 23°10'. Solution: Since this angle is given in Table IV, we find that log sin 23° 10' = 9.5948-10. Example 6-12. Find log cot 51°27'. Solution: From Table IV we obtain the values for the following tabulation: f log cot 51°20' = 9.9032 - 10 1 1 V \ \* 10' { { log cot 51°27' = 9.9032 - 10 - x J > 26 log cot 51°30' = 9.9006 - 10 The tabular difference is 26. Since 51°27' is ~ of the way from 51°20' to 51°30', x = 0.7(26) = 18.2, and we have log cot 51°27' = (9.9032-10) - .0018 = 9.9014-10. Example 6-13. Find the acute angle 6 if log tan = 9.7827-10. Solution: The positive part 9.7827 lies between the entries 9.7816 and 9.7845 in Table IV. The procedure for finding 6 may be indicated as follows: log tan 31°10' = 9.7816 - 10 11 10' x 11 Hence, j^ = ^9 , and log tan 31°(10 + x)' = 9.7827 - 10 log tan 31°20' = 9.7845 - 10 8 = 31°10' + 55UO') = 31°14'. ^ 29 Sec. 6-9 Right Triangles and Vectors 133 EXERCISE 6-4 In each of the problems from 1 to 15, find the value of the given logarithm. 1. log sin 48°20'. 2. log sin 21°20'. 3. log cos 86°20'. 4. log tan 88°30'. 5. log cot 10°20'. 6. log sec 43°50'. 7. log sin 13°26'. 8, log sec 48°57'. 9. log tan 41°14'. 10. log esc 78°32'. 11. log cot 68°43'. 12. log cos 18°18'. 13. log esc 83°16'. 14. log cos 1°8'. 15. log tan 51°34'. In each of the problems from 16 to 35, find the angle (or angles) between 0° and 360°. 16. log sin = 8.8059-10. 17. log tan = 8.3660-10, 18. log cos = 9.9959-10. 19. log sin = 9.1697-10. 20. log sec = 0.4625. 21. log cot = 0.4882. 22. log tan = 9.8483-10. 23. log tan = 0.1430. 24. log esc = 0.4081. 25. log sec = 0.3586. 26. log sin = 9.9567-10. 27. log tan = 9.7648-10. 28. log cos = 9.9755-10. 29. log cot = 9.8666-10. 30. log sec = 0.1967. 31. log esc = 0.3370. 32. log cos = 9.1860-10. 33. log cot = 1.5976. 34. log sin = 9.9974-10. 35. log sec = 0.3870. 6-9. LOGARITHMIC SOLUTION OF RIGHT TRIANGLES The solution of a right triangle by means of logarithms is exactly the same as by natural functions, except that for the actual numerical computation a table of logarithms of the natural functions is used in conjunction with a table of logarithms of numbers. The following example will illus- trate the procedure. Example 6-14. Solve the right triangle ABC, in which A = 21°40' and b = 8.43. Solution: The values of the known parts are indicated in Fig. 6-19. We see that B = 90° - 21°40' = 68°20'. To find the side a, we have tan21«40'=^, Hence, a = 8.43 tan 21°40', or log a = log 8.43 + log tan 21°40'. Arrange the work as follows: log 8.43 = 0.9258 (Table III) . log tan 21°40' = 9.5991-10 (Table IV) log a = 10.5249-10 Therefore, a = 3.35. (Table III) To find side c, we have — = cos 21°40\ c 1 34 Right Triangles and Vectors Sec. 6-9 Hence, ^ cos 21°40' and log c = log 8.43 - log cos 21°40'. Arrange the work as follows: log 8.43 = 10.9258-10 (Table III) log cos 21°40' = 9.9682-10 (Table IV) log c = 0.9576 Therefore, c = 9.07. (Table III) EXERCISE 6-5 In each of the problems from 1 to 8, solve the given right triangle. I. b = 100, A = 31°. 2. c = 3.45, a = 1.76. 3. A = 25°20', a = 63.4. 4. A = 88°17', c = 108.1. 5. c = 6.275, 5 = 18°45'. 6- a = 645.3, 6 = 396.3. 7. J5 = 27°9', a = 36.13. 8. 6 = 98.34, B = 18°48'. 9. If a railroad track rises 30 feet in a horizontal distance of one mile, find the angle of inclination of the track. 10. A force of 341 pounds and another force of 427 pounds act at right angles to each other. Find the magnitude of the resultant force and the angle it makes with each of the forces. II. A force of 628 pounds acts at 180°, and a force of 237 pounds acts at 270°. Find the direction and magnitude of the resultant. 12. An airplane is flying due east at a speed of 485 miles per hour, and the wind is blowing due south at 33.6 miles per hour. Find the direction and speed of the phane. 13. The westward and northward components of the velocity of an airplane are 363 and 487 miles, respectively. Find the direction and speed of the airplane. 14. The eastward and southward components of the velocity of a ship are 10.4 and 16.8 knots, respectively. Find the speed of the ship and the direction in which it is moving. 15. A force of 2673 pounds is acting at an angle of 47° 13' with the horizontal. Find its horizontal and vertical components. 16. A force of 162.4 pounds is just sufficient to keep a block at rest on a smooth inclined plane. If the block weighs* 783.1 pounds, find the angle at which the plane is inclined to the horizontal. 17. Two tangents are drawn from a point P to a circle whose radius is 14.32 inches. If the angle between the tangents is 32°28', how long is each tangent segment? 18. Two buildings of the same height are 11,640 feet apart. When an airplane is 8,000 feet above one of them, what is the angle of depression to the other one? 19. A cable which can withstand a pull of 10,000 pounds is used to pull loaded trucks up a ramp. If the angle of inclination of the ramp is 36°16', find the weight of the heaviest truck which can be safely pulled up the ramp with the cable. 20. The angle of elevation from one point on level ground to the top of a flagpole is 45°28'. From a point on the ground 25 feet farther away the angle is 39°56'. How high is the pole? 7 Trigonometric Functions of Sums and Differences 7-1. DERIVATION OF THE ADDITION FORMULAS Heretofore, we were concerned with relationships between trig- onometric functions of a single angle. We shall now establish certain fundamental identities involving two angles, in terms of the functions of the single angles. The following identities express functions of the sum and difference of two angles in terms* 6f the functions of the separate angles. (7-1) (7-2) (7-3) (7-4) (7-5) (7-6) sin (a + /3) = sin a cos /3 + cos a sin /3, cos (a + /3) = cos a cos /3 — sin a sin /3, sin (a — ]8) = sin a cos /3 — cos a sin j8, cos (a — 13) = cos a cos /3 + sin a sin /3, tan a + tan /3 tan (« + ^) = 1 _ tanatan ^' , . tan oj — tan j3 tan (a - j8) = — - tan a tan /3 We shall now prove the formulas for the sine and cosine by using the derivation developed by E. J. McShane. 1 +x Fig. 7-1. Fig. 7-2. 1 E. J. McShane. "The Addition Formulas for the Sine and Cosine," American Mathematical- Monthly, Vol. 48 (1941), pp. 688-89. 135 136 Trigonometric Functions of Sums & Differences See. 7-1 iiK Ip [cos [<x - /3), sin (a- 0)] / V* #\« \ ) ^ * X O ^HJp pa.o) Fig. 7-3. Let )3 and a be any two angles with the same initial side OW, as shown in Fig. 7-1. On their terminal sides we choose points P and Q, respectively, each at unit dis- tance from 0. Let d represent the distance from P to Q. We shall now make two computations for d 2 , using first OW, and then OP, as the #-axis. When OW is used as the #-axis of a coordinate system, as shown in Fig. 7-2, we find that the coordi- nates of P and Q are (cos /3, sin ft) and (cos a, sin a), respectively. Hence, by the distance formula, d 2 = (cos a — cos /3) 2 + (sin a — sin /3) 2 = cos 2 a — 2 cos a cos /3 + cos 2 /3 + sin 2 a — 2 sin a sin (3 + sin 2 /3. Since cos 2 a + sin 2 a = cos 2 /3 + sin 2 /3 = 1 , d 2 = 2 — 2(cos a cos /3 + sin a sin /3). Let us now use OP as the a?-axis, as shown in Fig. 7-3. Then the coordinates of P are (1,0) and those of Q are [cos (a-/3), sin (a — /3) ] . Hence, d 2 = [cos (a - jg) - l] 2 + sin 2 (a - 0) = cos 2 (a - 0) - 2 cos (a - j8) + 1 + sin 2 (a - 0) = 2 - 2 cos (a - j8). Equating the two expressions for d 2 yields 2 — 2(cos a cos + sin a sin j8) =2 — 2 cos (a — |8). Therefore, (7-4) cos (a — 0) = cos a cos + sin a sin 0. This establishes (7-4). Setting a = 90° in (7-4) , we find that (7-7) cos (90° - 0) = sin 0. If in (7-7) we let /3 = 90° - y, we have (7-8) sin (90° - 7) = cos 7. From (7-7), (7-8), and (7-4), we obtain sin (a + 0) = cos [90° - (a + 0)] = cos [(90° - <*) - 0] = cos (90° - a) cos + sin (90° - a) sin 0, Sec. 7-1 Trigonometric Functions of Sums & Differences 137 or (7-1) sin (a + ft) = sin a cos ft + cos a sin ft. This establishes (7-1). Since cos (-ft) = cos ft and sin (-ft) = -sin ft, we have, as a con- sequence of (7-4), cos (a + ft) = cos [a - ( — j8)] = cos a cos (— ft) + sin a sin (— ft) or (7-2) cos (a + ft) = cos a cos ft — sin a sin j8. Similarly, from (7-1) it follows that (7-3) sin (a — ft) = sin a cos (— j8) + cos a sin ( — ft) = sin a cos ft — cos a sin ft. To prove (7-5), we use the relationship tan 6 = s and (7-1) COS and (7-2). We then obtain , , ns sin (a + ft) sin a cos + cos a sin tan (a + ft) = r~~7oT ~ 5 : : — 5 * cos (a + ft) cos a cos p — sin a sin p Dividing each term of the numerator and denominator by cos a cos ft, we have sin a cos cos a sin co s a cos cos a cos ft ___ tan a + tan ft ^ cos a cos ft sin a sin ft 1 — tan a tan ft xx cos a cos ft cos a cos ft Hence, , , . o ,*, rN / , /on tan a + tan p (7-5) ton < a + fl = i- -ton«tanfl - We may obtain (7-6) in a similar manner from (7-3) and (7-4), or from (7-5). Example 7-L Find the exact value of sin 75°. Solution: Substituting 45° for a and 30° for in (7-1), we obtain sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° _\/2 Vj \/2 l t ~ 2 ' 2 + 2 V Therefore, ,^ sin 75° = ^ (\/3 + 1). 138 Trigonometric Functions of Sums & Differences Sec. 7—1 Example 7-2. Find the exact value of tan (a 4- P) if a is a second-quadrant 3 5 angle such that sin a = ■=■ , and P is a third-quadrant angle tan P = — • Solution: Let a be a second-quadrant angle for which y = 3 and r = 5. 3 Hence, x = — 4. It follows that tan a = - - • Using (7-5), we have _§+£ * / , a\ 4 1 2 16 tan (a + 0) = T-gTy^T = ~ 53 ' -(-)(&) EXERCISE 7-1 In each of the problems from 1 to 8, find the exact value of the given function. 1. cos 75°. 2. tan 105°. 3. sin 135°. 4. sin 15°. 5. tan 195°. 6. cos 195°. 7. tan 15°. 8. cos 105°. 9. If a is a third-quadrant angle and p is a second-quadrant angle, and 3 5 sin a = - -= and cos P = -rr> find sin (a + 0), cos (a -f P), and tan (a + P). 10. If tan a = ^ and a - 6 = 45°, find tan 0. 11. If tan a = 3 and a + p = 180°, find tan p. 3 1 12. If tan a = j and tan = - > find sin (a + 0) and cos (a +0), where a and are first-quadrant angles. 13. If cos a = - and cos P = *= > find sin (a — 0) and cos (a — 0), where a and are acute angles. If sin a = -= » tai o and cos (a -f- j8). „ . 3 If sin a. = ■=• > 5 cos (a — p). 4 5 14. If sin a = ■= » tan = — — > and a and are both obtuse, find sin (a + P) t+Ph 3 24 15. If sin a = ■= > cos P = jr? > a is obtuse, and is acute, find sin (a — p) and 2 1 16. If cos a = - - > sin = - > and a and are obtuse, find sin (a + j8) and cos (a + P). 3 5 17. If sin a = - > cos = — j$ > a is obtuse, and P is in the third quadrant, find tan (a + P) and tan (a - p). Prove each of the following identities: a/2 18. sin (a - 45°) = -5- (sin a - cos a). 19. cos (a - 7r) = - cos a. 20. sin (a - 0) _ / tt\ 1 + tan a — : r— —• = cot — cot a. 21. tan I a + -t ) = ^ 7 * sin a sin j3 \ 4 / 1 — tan a 22. cos 2 a = cos 2 a - sin 2 a = 2 cos 2 a - 1 = 1 - 2 sin 2 a. Sec. 7-2 Trigonometric Functions of Sums & Differences 1 39 OO • O O • OA Sil1 2 a COS 2 Q! 23. sin 2 a = 2 sin a cos a. 24. — = sec a. sin a cos a 25. sin ( a + fl) + sin (a - g) __ sin (a + 0) _ , , . p — __ tan a< 26. ^ = tan a -f tan 0. cos (a + p) + cos (a - p) cos a cos 27 s i n Cg + 0) _ tan g -f tan ft ^ ^ cos (a - g) __ 1 + tan a tan |8 ^ sin (a - P) tan a - tan cos Ja -f jS) ~~ 1 - tan a tan oq cos (g - g) _ 1 + tan a ta n cot a cot - 1 29 ' ^(alj) " la^TTlanT ' "' COt ( " + ft = ^ot^T"c"oT^ " 31. Bin (a 4-0) sin (a -0) = ^ ^ _ ^ cos 2 a cos 2 p 32. sin (A + B + C) =sin A cos £ cos C + cos A sin B cos C -f cos A cos J3 sin C - sin A sin B sin C. 33. sin 3 a = sin 5 a cos 2 a — cos 5 a sin 2 a. 34. cos (a — 0) cos (a + 0) = cos 2 a — sin 2 = cos 2 /3 — sin 2 a. 35. sin 2 ( a + t) "~ cos2 ( a + t) ~ sin 2 a> 7-2. THE DOUBLE-ANGLE FORMULAS If we let fS ~ a in (7-1), (7-2), and (7-5), we obtain functions of twice a given angle in terms of the functions of the angle itself. Thus, we have the following identities : (7-9) sin 2 a = 2 sin a cos a, (7-10) cos 2 a = cos 2 a — sin 2 a, and t» t i \ o 2 tan a (7-11) tan 2 a = 1 — tan 2 a We may obtain two other useful forms for cos 2 a from (7-10) by using in turn cos 2 a — 1 — sin 2 a and sin 2 a = 1 — cos 2 a. These forms are given by the identities (7-12) cos 2 a = 1 - 2 sin 2 a, and (7-13) cos 2 Q! = 2 cos 2 a - 1. The following illustrations give an indication of the possible applications of (7-9), (7-10), and (7-11). The student should study them carefully. sin 4 a = sin 2 (2 a) = 2 sin 2 a cos 2 a, /a\ . ot ot sin a = sin 2 I ~ ) = 2 sin ^ cos ^ > cos - = cos 2 1^1 = cos 2 - — sur ^ • 1 40 Trigonometric Functions of Sums & Differences Sec. 7—2 Example 7-3. Find the exact value of sin 120° by means of a double-angle formula. Solution: We use (7-9) to obtain sin 120° = sin 2(60°) = 2 sin 60° cos 60° =»(#G)-# Example 7-4. Derive a formula for cos 3 a in terms of cos a. Solution: Applying the identity for cos (a -f 0), and the double-angle formulas, we have cos 3 a = cos (a -f 2 a) = cos a cos 2a — sin a sin 2 a = cos a (2 cos 2 a - 1) - sin a (2 sin a cos a) = 2 cos 3 a — cos a — 2 sin 2 a cos a = 2 cos a (cos 2 a — sin 2 a) — cos a = 2 cos a (2 cos 2 a — 1) — cos a = 4 cos 3 a — 3 cos a. ^ i « ^ t» ., . , ...sin 3 cos 3 n Example 7-5. Prove the identity — r-^ -r- — 2. r sin cos Solution: First combine the fractions on the left side and then reduce the result to the right side. Thus, sin 3 __ cos 3 __ sin 3 cos - cos 3 sin _ sin (3 0-0) sin cos ~~ sin cos "~ sin cos sin 2 2 sin 2 sin cos sin 2 = 2. 7-3. THE HALF-ANGLE FORMULAS Functions of an angle in terms of the functions of twice that angle can be obtained directly from (7-12) and (7-13) . If cos 2 a = 1 — 2 sin 2 a is solved for sin a, we obtain V 1 cos 2 a sin a Also, by solving cos 2 a = 2 cos 2 a — 1 for cos a, we have ;a = ±|A + cos 2 a cos r " * " 2 Since these formulas may be equally well regarded as expressing functions of half an angle in terms of the functions of the given angle itself, the same relationship is retained if the identities are written (7-14) and (7-15) . a sin ^ = ±l />- cos a 2 a cos 5 = ±l /- ■ cos a 2 Sec. 7-3 Trigonometric Functions of Sums & Differences 141 These are the so-called half -angle formulas for the sine and cosine. From (7-14) and (7-15) we obtain, by division, tr, -,n\ a si n a/2 , A /T < 7 - 16 > tan 2 = ^572 = ± yr — cos a t/2 y 1 + cos a The algebraic signs in (7-14), (7-15), and (7-16) are deter- mined by the quadrant of a/2. If we rationalize, in turn, the numerator and the denominator of the right hand side of (7-16) , we obtain tan %=±n — POS 2 cos* a (1 + cos a) 2 v- (1 + cos a) 2 or (7-17) Similarly a sm a tan ^ = j—. 2 1 + cos a tan f = ± \/ { \ ~ ™f = ± |/E^! 2 V 1 — cos 2 a f snr a or (7-18) tan a 1 — cos a sin a The student will note that, in deriv- ing (7-17) and (7-18), we have dropped the ± sign in each formula. The validity of this step should be verified by consideration of the signs of tan a/2, sin a, and 1 ± cos a. Thus, tan a/2 and sin a necessarily have the same sign, while 1 =t cos a is non- negative. Example 7-6. Find the exact value of tan 22.5°. r-24 Fig. 7-4. Solution: Since the exact values of the functions of 45° are known, we may use (7-16), (7-17) or (7-18). Selecting (7-18), we obtain * ™r Q j. 45° 1 - cos 45° 2 tan 22.5° = tan — = — r-— — = 7T~ 2 sm 45° y/2 V2 1 42 Trigonometric Functions of Sums & Differences Sec. 7—3 24 Example 7-7. Given tan 2a= — =- > where 2 a is a second-quadrant angle. Find sin a and cos a. 24 Solution: Since 2 a is an angle whose tangent is — =- » we may find a point on the terminal side of the angle with x = — 7 and y = 24, as shown in Fig. 7-4. Thus, r = V49 + 576 = 25. Therefore, a/1 +7 '25 4 and COS Of = /l/ — -7/25 3 2 5' a Example 7-8. Given that tan - = u, find sin a and cos a: in terms of w. Solution: Squaring both sides of (7-16), we have n a 1 — cos a tan 2 - = — • 2 1 + cos a ol 1 " — cos a. Substituting u for tan - > we get u 2 = ■= — ■ • If we solve this equation for * \j 4.w 2 1 4- cos a cos a, we rind that 1 - u 2 cos a = — — - • 1 -f u 2 If we substitute this value of cos a in the relationship sin 2 a + cos 2 a = 1, we obtain i/] 7l - fc 2 \ 2 2?< Note that we can drop the ± sign, just as we did in (7-17) and (7-18), since tan a/2 and sin a always have the same sign. EXERCISE 7-2 In each of the problems from 1 to 8, find the exact functional value by using an appropriate double-angle or half-angle formula. I. sin 22.5°. 2. cos 15°. 3. sin 120°. 4. cos 90°. 5. sin 67.5°. 6. cos 67.5°. 7. tan 67.5°. 8. tan 60°. 12 9. It is known that cos 6 — — ^ and 6 is positive in the second quadrant. Find : lo a. sin 20. b. cos 6/2. c. tan 26. d. cot 26. e. sin 6/2. f. cos 26. g. sin 36. h. tan 40. 40 10. It is known that sin 6 = — — and 6 is positive in the third quadrant. Find : a. sin 26. b. cos 6/2. c. tan 26. d. cot 26. e. sin 6/2. f. cos 26. g. sin 30. h. tan 40. II. It is known that tan = - 3 and is positive in the fourth quadrant. Find: a. sin 20. b. cos 0/2. c. tan 20. d. cot 20. e. sin 0/2. f. cos 20. g. sin 30. h. tan 40. Sec. 7-4 Trigonometric Functions of Sums & Differences 1 43 In each of the problems from 12 to 28, write the given expression in terms of a single function of a multiple of 0. Make use of appropriate formulas to reduce the answer to as few terms as possible. 12. 2 sin ^ cos | • 13. cos* ~ - sin* j . 14. 2 cos* 30 - 1. 2 tan 30 1C , .50 t _ . nik A /l - cos 40 1 - tan* 30 ' 16 ' * " Sin2 T ' 17 - Sin 2$ V 2 — *V. m ___. „ .._. /, 9n sin 20 COS 20 ZU. : jr — n * sm cos 18. (sin - - cos 2) • 19- cos 4 - sin 4 0. 21 2 00 2cot0 ^ 1 - tan* 0/2 cot - tan 1 + cot* 1 + tan* 0/2 ' 04 4. Q , * ^ ok sin 0/ 2 o* 2 tan 0/2 24. cot x + tan 5 • 25. - '-^r • 26. t— — — -~-rr • 2 2 1 - cos 0/2 1 4- tan* 0/2 27. j 7 C0S H • 28. cos* (0*) - sin* (0*). 1 + cos 30 Prove each of the following identities: 29. sin 40-4 cos 0(sin 0-2 sin 3 0). 30. cos 40 = 8 cos 4 0-8 cos* + 1. 31. sin 20 = (1 + cos 20) tan 0. 32. 1 + sin 20 = (sin + cos 0) 2 . 00 1 - tan* 0/2 2(1 - cos 0) ***-*•* ■ a « <>* 33. £-*— = . — — - • 34. 4 sin cos* = sin + sin 30. cos sin* 7-4. PRODUCTS OF TWO FUNCTIONS EXPRESSED AS SUMS, AND SUMS EXPRESSED AS PRODUCTS By adding and subtracting corresponding members of (7-1), (7-2), (7-3), and (7-4), we obtain (7-19) sin (a + 0) + sin (a - j8) = 2 sin a cos 0, (7-20) sin (a + |8) - sin (a - 0) = 2 cos a sin 0, (7-21) cos (a + 0) + cos (a - 0) = 2 cos a cos 0, (7-22) cos (a + 0) - cos (a - 0) = - 2 sin a sin 0. If we reverse these identities, they become the following product formulas, which express given products of sines and cosines as sums or differences : (7-23) sin a cos = ^ [sin (a + 0) + sin (a - 0)], (7-24) cos a sin = 2 [sin (a + 0) - sin (a - 0)], (7-25) cos a cos = 2 [cos (a + 0) + cos (a - 0)], (7-26) sin a sin = - | [cos (a + 0) - cos (a - 0)]. 1 44 Trigonometric Functions of Sums & Differences Sec. 7-4 To obtain the sum formulas, which express given sums or differ- ences of sines and cosines as products, we first let a + /3 = x and a — (3 = y. Then, solving for a and /3, we have % + y j q x — y a = — p^- and j8 = — ^ • Substituting these values of a and /3 in (7-19), (7-20), (7-21), and (7-22), we obtain the sum formulas. These are (7-27) sin x + sin y = 2 sin (^-y^) cos (^-y^) > (7-28) sin a: - sin 2/ = 2 cos (^-y^) sin ( - ~ y J > (7-29) cos x + cos y = 2 cos (^y^) cos (^-y^) > (7-30) cos x - cos y = - 2 sin (^jpQ sin (^y^) • Example 7-9. Express sin 3 a cos 5 a as a sum of sines. Solution: Using (7-24) and replacing a by 5 a and by 3 a, we obtain cos 5a sin 3a = r[sin (5a + 3a) - sin (5a - 3a)] = -[sin 8a - sin 2a]. Example 7-10. Express cos 40 + cos 20 as a product of cosines. Solution: Using (7-29) and replacing x by 40 and y by 20, we have cos 40 -f cos 20 = 2 cos — - — cos — - — = 2 cos 30 cos 0. ^i i ~ *i t^ .li • i x-x sin 7x — sin 5x , Example 7-11. Prove the identity = — — — — r- = tan x. COS (X ~j~ COS ox Solution: / 7x + 5x \ . / 7x - 5x \ sin 7x - sin 5x _ V 2 / V 2 / (Ix + 5z\ (Ix - 5a:\ 2 cos (— y— j cos (— y-J 2 cos 6a; sin a: cos 7.t -I- cos 5a: n /7x + 5x\ (7x - bx n _ # _ i cog i - tan x. 2 cos 6a; cos x EXERCISE 7-3 In each of the problems from 1 to 10, write the given expression as a sum or difference of two sines or two cosines. 1. sin 30 cos 40. 2. 2 sin 40 cos 20. 3. 2 sin 60 cos 40. 4. sin cos 40. 5. cos 40 cos 20. 6. 2 sin 65° cos 15°. 7. sin 28° sin 20°. 8. cos 21° cos 31°. 9. sin 50 sin 0. 10. sin 110 sin 30. Sec. 7-4 Trigonometric Functions of Sums & Differences 1 45 In each of the problems from 11 to 20, write the given expression as a product of sines and cosines. Hint: In problems 18, 19, and 20, note that cos = sin (90° - 0). 11. sin 30 + sin 20. 12. cos — cos 40. 13. sin 60 + sin 30. 14. sin 40° + sin 20°. 15. cos 80° - cos 20°. 16. sin 30° - sin 80°. 17. sin 40° + sin 25°. 18. sin 64° + cos 38°. 19. sin 40° + cos 44°. 20. sin 65° - cos 33°. Prove each of the following identities: 21. sin + cos = y/2 cos (d - -j) • 22. x , x a cos (a — 0) tan a + cot = — r~^ • cos a gin p no cos0 , sin0 OQ 23. 72 + Ta = cos 30. sec 40 esc 40 24. sin , cos0 . ^ sec 30 esc 30 25. Sin2 9 ! l + ^ta* cos 20 + cos 40 26. sin 29 - sin A cos 30 + cos 27>8 in«-f S in^ =tan a+0. cos a + cos p 2 sin (# + t) + sin (* ~ t) = ^ Sin < 29. sin (0 + -| ) - sin (0 - ~) = \/3 cos 0. cos(-| +0) - cos(-| - 0) = - V3sin0. 30. , sin - cos + 1 , 31. . 2 « ITTT = tan 7r • sm + cos + 1 2 32. sin (a + j8) sin (a - j8) = sin 2 a - sin 2 j8 = cos 2 fi - cos 2 a. 33. cos (a + fl) cos (a - jS) = cos 2 a - sin 2 j8 = cos 2 - sin 2 a. Graphs of Trigonometric O Functions; Inverse Functions and Their Graphs (-1,0) 8-1. VARIATION OF THE TRIGONOMETRIC FUNCTIONS In Section 3-2, the trigonometric functions were defined in terms of the coordinates (x, y) of the point P(t), where the number t represents the directed length of the arc of a unit circle measured from the point (1,0). Later, in Section 3-9, an equivalent defini- tion was given in terms of an angle 9 in standard position. We shall now consider the variation of the trigonometric functions in the four quadrants as t, and with it 0, increases from to 2n. From Fig. 8-1, we can read off the vari- ations shown in Table 8-1. For example, by noticing the changes in y as t increases con- tinuously from to 27T, we find that sin t varies from to 1 in the first quadrant, from 1 to in the second, from to — 1 in the third, and from -1 to in the fourth. Similar considerations lead to the results for the cosine and tangent. Recalling that esc t = — - > we know that if either of these func- sin I tions increases the other decreases. Hence, the variation in esc t can be determined from the variation in sin t. Similarly, we may learn about the variation of sec t from that of cos t, and about the variation of cot t from that of tan t. We found in Example 3-2 that tan n/2 is undefined, which means that tan t has no value when t = 7r/2. For the sake of easier tabu- 146 Fig. 8-1. See. 8-2 Trigonometric Functions; Inverse Functions TABLE 8-1 Variation of Trigonometric Functions 147 From To From To From To From To t tt/2 w/2 T T 3ir/2 3tt/2 2ir sin t 1 1 -1 -1 cos t 1 -1 -1 1 tan t CO — 00 00 — 00 CSC t 00 1 1 00 — 00 -1 -1 — CO sec t 1 CO — 00 -1 -1 — 00 00 1 cot t 00 — 00 00 — CO lation of this result in Table 8-1 we have employed the much used symbols oo (infinity) and -co. These symbols merely signify that in the neighborhood of tt/2 or one of its odd multiples the value of tan t is very large numerically. They are not to be used as numbers. 8-2. THE GRAPH OF THE SINE FUNCTION To construct a graph representing the variation of the sine, we let x denote a real number or the value of an angle measured either in radians or in degrees, and we let y denote the corresponding value of the function. Corresponding values of x and y are plotted +x Fig. 8-2. 148 Trigonometric Functions; Inverse Functions Sec. 8-2 as points on a rectangular coordinate system. We can infer the general appearance of the curve from the results summarized in the preceding section, but an exact representation is more readily obtained by using a table of sines. Let us now construct the graph of y = sin x from x = -tt/2 to x = 2n. The following values, found by the methods of Section 3-2, are used to obtain the curve in Fig. 8-2. X 7T " 2 T 3 7T " 6 "6 7T 7T 2~ 2tt 3 5?r 6 y -1 -0.87 -0.5 0.5 0.87 1 0.87 0.5 X T 7tt 6 4w 3tt 2 5tt 3 llTT 6 2tt y -0.5 -0.87 -1 -0.87 -0.5 While the choice of a scale is arbitrary, a better propor- tioned graph results if the same unit of length is used on both axes. The unit so selected will represent the number 1 on the y-axis and one radian on the #-axis. In terms of this unit a suitable length can then be marked off on the #-axis to represent 2tt or 360°. 8-3. THE GRAPHS OF THE COSINE AND TANGENT FUNCTIONS Using the table of trigonometric functions, the student should make a table of corresponding values for y = cos x and one for *»x y - cos x Fig. S-3. Sec. 8-4 Trigonometric Functions; Inverse Functions 149 7T/2 y - tan x 37T/2 27T +»x Fig. 8-4. y = tan x, similar to that used for y = sin x in Section 8-2. Study the graphs in Fig. 8-3 and Fig. 8-4 on the basis of the tables you have made. If we compare Fig. 8-3 with Fig. 8-2, we see that the graph of y = cos x may be obtained from the graph of y = sin x by moving the graph of y = sin x to the left a distance of tt/2 units. This fact can be checked by using the relationship cos x = sin (x + ir/2) 9 from which it follows that cos = sin 7r/2, cos tt/6 = sin 27r/3, and so on. 8-4. PERIODICITY, AMPLITUDE, AND PHASE The trigonometric functions are among the simplest of a large class of functions which are periodic. As a preliminary to defining periodic functions, we shall call attention to some examples of phenomena which recur periodically, such as the rotation of the earth about its axis, sound and water waves, the vibration of a spring, and many other vibratory and wavelike phenomena. The behavior of the object involved in a phenomenon of periodic nature determines the type of function that is required to represent it properly. We note particularly that, because of the recurrence characteristic of such a phenomenon, the values which the function assumes in any interval of given length are also taken on in any other interval of the same length. This statement apparently indi- cates that a function of x is periodic with period p if, for every value of x, the function returns to the same value when x is increased by p. More specifically, we state the following. 1 50 Trigonometric Functions; Inverse Functions Sec. 8-4 Definition. A function f(x) is said to be periodic if there is a non- zero number p for which f(x + p) =f(x) for all numbers x in the domain of f(x). Any such number p is called a period; the smallest positive number p satisfying the requirement is called the period. Evidently, if p is the period of f(x), then np is a period, for every integer n. Periodicity of Trigonometric Functions. No matter which of the two viewpoints is considered in the definition of the trigonometric functions, we shall see that, if g t ^ 2n f then any trigonometric function of (t + 2tt) = same function of t. According to the definitions given in Sections 3-1 and 3-2, P(t) and P(t + 2Tr) represent the same point on the circumference of the unit circle. To locate P(t) we start at (1,0) and proceed around the circle in the proper direction a distance of |£| units. To locate P(t + 2ir) we continue another 2tt units from the point P(t) . This merely adds another complete revolution, and we arrive at the same point P(t). If we consider the definitions of the functions in terms of angles, as given in Section 3-9, we note that the angle + 2n is coterminal with and that any trigonometric function has the same value for coterminal angles. Period of Sine, Cosine, Cosecant, and Secant. From a study of the sine curve, it is apparent that sin x assumes all values between —1 and +1 as x, starting from any value, varies through 2tt units. In other words, the graph of the function repeats itself during each interval of length 2tt, for positive and negative values of x. Or stated more concisely, sin (x + 27r) = sin x. This is equivalent to saying that sin x is a periodic function of x, and that 2tt is a period. It remains now to show that 2tt or 360° is the smallest positive number p for which sin (x + p) = sin x and for which cos (x + p) = cos x. Since, by definition of a period p, sin (x + p) = sin x for any value x, we shall select for the purpose of our proof the particular value x = it/2. We then have . /7T .• \ . 7T sin (^r + v) = S1 *i y = 1- But since sin ( -=- + pj = cos p } it follows that cos p = 1. Sec. 8-4 Trigonometric Functions; Inverse Functions 151 Hence, p must be an even multiple of tt. The smallest even multiple of 7r is 2tt, which must also be the smallest positive period of the sine function. In a similar manner, we find that 2n is also the period of the cosine function. Because of the reciprocal relationships existing between the sine and cosecant and between the cosine and secant, 2tt is also the period of the cosecant and the secant. Period of Tangent and Cotangent, To find the period of the tangent, we write tan (x + V) = tan x } and we let x = 0. Then tan p = 0, and we find that tt is the period of the tangent. A similar argument shows that tt is also the period of the cotangent. Period of sin bx. We have just seen that the period of sin x is 2tt. We shall now determine the period of sin bx, where 6 is a positive constant. That is, we want to know the smallest positive change in x which will produce a change of 2ir in bx. If p repre- sents this change in x, we can find p from the relationship b(x + p) = bx + 27T. Solving for p, we immediately find that 2tt P = T - Thus, the period of sin bx is equal to the period of sin x divided by b, that is, 27r/&. Similarly, it can be shown that 2tt /b is also the period of cos bx, esc bx, and sec bx. It is also true that the period of tan bx and cot bx is n/b. 2> J 1 \ / \ y=*sit\2x 7T/2\ A 37r/2\ Att ■■"^ A. -1 1 \ y~*2 sinx \ / -2- Fig. 8-5. 152 Trigonometric Functions; Inverse Functions Sec. 8-4 Let us consider the graph ofy = sin 2x shown in Fig. 8-5. Since the period of sin 2x is 2ir/b = 27r/2 = 7r, the function will assume the same range of values in the interval from to tt that sin x takes in the interval from to 2tt. Note that the graph of y = 2 sin x is similar in form to that of y = sin x, which is shown in Fig. 8-2. The period is 2tt for both curves. However, for any given value of x, the corresponding value of y in y = 2 sin x is twice as large as is the corresponding value of y in y = sin x for the same value of x. In the graph ofy — a sin x, where a > 0, the greatest value of y is a, and the smallest value of y is —a. The constant a is called the amplitude of the function or of the graph. Thus, the amplitude of the graph of y = sin x is 1, while that of the graph of y = 2 sin x is 2. In general, for the function y = a sin x, where a is a real number, the amplitude is equal to |a| and the period is equal to 2tt. Also, in general, for the graph ofy — a sin bx> where a and b are real, the amplitude is \a\ and the period is 2rr/b. Phase Angle. Since the graph of y — cos x may be obtained from the graph ofy = sin x by shifting it to the left a distance equal to 7r/2 units, we say that the graph of y = cos x differs in phase by 7r/2 from the graph ofy — sin x. The amount of horizontal dis- placement of two congruent graphs, amounting to tt/2 radians in this case, is called the phase difference, or the phase angle. The amplitude and the period are the same for y = cos x as for y = sin #. *-x / » cos 2x - sin (2 .r + 7T/2) = sin 2 (.r + 7T/4) Fig. 8-6. Now consider the graph of y = cos 2# = sin 2(x + ~) in Fig. 8-6. This curve may evidently be obtained from that ofy = sin 2x in Fig. 8-5 by a shift of 7r/4 units to the left in the x direction. Hence, the graph of y = sin (2x + ~) differs in phase from that of y = sin 2x by 7r/4 radians. Sec. 8-4 Trigonometric Functions; Inverse Functions 1 53 A simple method for finding the phase displacement is to locate the point near the origin for which the function sin (2x+^) equals zero ; that is, to find the smallest numerical value of x that 7T TV makes the quantity x + - zero. We have then sin2(#-f-) =0 7T 4 4 when x + - = or when x = — tt/4. Hence, the phase difference is 7r/4 radians, and the shift of the graph is toward the left since the sign of x is negative. It can be shown that, in general, the phase displacement of any trigonometric function of (bx + c), where b > 0, is to the right or left by \c/b\ radians (or degrees). The direction of the displace- ment depends on whether c/b is negative or positive. Finally, we arrive at the conclusion that for the graph of y = a sin (bx + c) the amplitude is |a| and the period is 27r/b. Also, its phase differs from that of y = sin x by c/b. Because of its usefulness in many applications, we shall illustrate by means of an example a procedure for reducing an expression of the form A sin + B cos # to the form a sin (0 + a) . Example 8-1. Reduce A sin -f B cos to the form a sin (0 + «)• Solution: Write A sin + B cos = VA 2 + B 2 (- _ sin + — = cos d) • \\/^ 2 + B 2 VA 2 +B 2 / A B The absolute values of the coefficients — — and — =— cannot be greater VA 2 + £2 VA 2 + B 2 than 1, and the sum of their squares is 1. Hence, they may be taken as the cosine and sine, respectively, of some angle a. Therefore, the expression A sin 9 -f B cos $ becomes y/A 2 + B 2 (cos a sin 6 + sin a cos 6) = VA 2 + B 2 sin (0 + a), where A = a cos a and J5 = a sin a. Example 8-2. Express % = sin 20 - V3 cos 20 in the form y = a sin (60 + a), and sketch the graph. Solution: Since A = 1 and £ = - \/3, we have y/A 2 + B 2 =2. Therefore, a, sin 20 + V3 cos 20 = 2^ sin 20 - ^ cos 20^. If we identify this result with the expression \JA 2 -f B 2 (cos a sin 20 + sin a cos 20), we have cos a = ^ and sin a = — - • angle and may be taken equal to - w/3. 1 /*\ we have cos a = ^ and sin a = -~ • It follows that a is a fourth quadrant 154 Trigonometric Functions; Inverse Functions Sec. 8-4 We have, finally, - _ i / *■ \ • y = i(sin 20 - V3 cos 20) = ± sin (20 - j J • The amplitude of this function is 5 > its period is 7r, and its phase angle is tt/6. -1/2^: j--l/2sin(20-7T/3) Fig. 8-7. Since the interval from (x/6, 0) to (7tt/6, 0) is one period in length, the part of the curve obtained for this interval may be repeated indefinitely in both directions to give the complete curve. The curve is shown in Fig. 8-7. (Here for convenience we have employed different units of length on the two axes.) EXERCISE 8-1 In each of the problems from 1 to 24, find the period, amplitude, and phase angle of the trigonometric function. 2. sin ~ • 5. ^ sin j 6 • 8. Braz- il. 3 sin 56. 14. sec 96. 17. tan (tt 6 + 4). 19. esc (tt 6 + 7). 20. cot (2tt 6 - tt). 22. 6 - cos 40. 23. 5 + 3 sin (26 - j\ • 1. 3 sin 0. 4. C0S 2' 7. 4 tan 1 10. sin 7r0. 13. cot 60. 16. 3 esc 3x 3. 2 cos 0. 6. 2 cot 1 • 9. COS -j-0 • 4 12. 5 tan 7r0 15. | cot 40. 18. cos (30 - 2). 21. 2 + sin 0. 24. 4 + 2 cos (20 + -|) In each of the problems from 25 to 30, sketch the graph of the given function by constructing a table of values. 25. y = cos •= x • 26. y = tan 5 z • 27. 2/ = 2 sin 3.r. 28. 2/ = 5 cos 5 a: • 6' 29. y = 2 tan j x * 30. 2/ = 3 sin 2z. Sec. 8—5 Trigonometric Functions; Inverse Functions 1 55 Sketch each of the following graphs without constructing a table of values. 2 1 31. y = sin ^ x. 32. y = cos 4z. 33. y = tan ^ #. 34. y = 2 cos 3.r. 35. y = 3 cos to. 36. */ = ^ tan ^ Xt 11 14 / ir\ 37. 2/ = o sin o Xm 38. y =z cos ^ Z. 39. y = 5 cos ^4x + y J 40. ?/ = cos (x + 2). 41. y = sin L? ~ s) • 42. y = cos (3z - 2). 43. y = sin (2x + 1). 44. ?/ = cos (27rx - ir). 45. 2/ = sin f j x + 7 J • 46. ?/ = sin + cos 0. 47. y = sin — cos 0. - - 13 48. y = V3 sin 20 + V5 cos 20. 49. y = 5 cos tt0 - - sin tt0. 50. y = V2 cos 30-3 sin 30. 51. y = 2 sin - cos 20. 8-5. INVERSE FUNCTIONS In Section 2-3, we defined a function by setting up a rule of correspondence between twa sets of numbers, X and Y, called the domain of definition of the function and the range set of the func- tion, respectively. The function was called single-valued if just one number y of the set Y is assigned to each number x of the set X. If more than one number of Y corresponds to some value of x, the function is multiple-valued. If we know that y = f(x), we may pose a reverse problem. We assume y to be given and ask for all corresponding values of x. Naturally, y is limited to lie in the range of the given function, since otherwise no x exists. The function which makes correspond to each such y all values of x for which y = f(x) is called the inverse function corresponding to the given function. We shall begin our discussion of inverse functions with an example in which X is the set of all real numbers, Y is the set of all non-negative real numbers, and the correspondence is deter- mined by the relationship y = x 2 . Ordinarily, we assign values to x in order to calculate values of x 2 . In this case, we have a rule of correspondence that assigns just one number y to ea$h chosen number x. Hence, y is a single-valued function of x. The s graph ofy = x 2 is shown in Fig. 8-8. Assume now that y is given and that we wish to determine cor- responding values of x. To do this we solve the given equation x 2 = y for x, and obtain two numbers x = y/y and* x = —y/y cor- responding to every. non-negative number y. Since the rule of 156 Trigonometric Functions; Inverse Functions Sec. 8-5 correspondence assigns two values of x to each chosen number y, we see that # is a double-valued function of y. In this case, the admissible values of y are restricted to zero and the positive real numbers, while those of x comprise all real numbers, as was indi- cated in the specification for the sets X and Y. i 1 I i >*x H 1 1 1 1 1 1 — I >~X - V 2 x~y Fig. 8-8. Fig. 8-9. In the study of mathematics, we generally prefer to use the sym- bol x to represent the independent variable and y to represent the dependent variable. To be consistent with this preference, we shall call y = x 2 and y = ±\/x inverse functions, each being called the inverse of the other. The graph of y = ±\/x is obtained by plotting that of x = y 2 , as shown in Fig. 8-9. We note that the roles of x and y are interchanged in the two equations y=x 2 and x = y 2 . Thus, we see that the curve in Fig. 8-9 is actually the curve of Fig. 8-8 with the axes interchanged and one of them reversed in direction. 8-6. INVERSES OF THE TRIGONOMETRIC FUNCTIONS The Inverse Sine. Let us consider the function y = sin x and attempt to apply a discussion similar to that in Section 8-5. Here, as has been noted, the domain is the set of all real numbers, while the range is the interval -1 S j/ ^ 1. Referring to the graph of y = sin x in Fig. 8-2, and recalling the periodic properties of this graph, we see that, for every given number y such that — 1 :sS y ^ 1, there are infinitely many values of x such that y = sin x. To desig- nate the totality of all values of x such that y = sin x, we write -i x = sm~ y, which is read x is the inverse sine of y. The student should note care- fully that the symbol sin* 1 y must be distinguished from (sin y)-\ which equals - — or esc y. sin v Sec. 8-6 Trigonometric Functions; Inverse Functions 1 57 Another notation that is frequently used to represent this inverse function is x = arc sin y, which is read x is the arc sine of y. In order to conform to the preferred practice of considering y as a function of x, we may designate the inverse of the sine func- tion by writing y = sin™ 1 x or y = arc sin x. We note the following properties of this inverse function. If # is a number such that \x\ > 1, then y does not exist. This property follows from the fact that the sine function takes on only the values from — 1 to 1. Hence, the inverse sine function sin -1 x is defined only when — 1 ^ x ^ 1. For example, sin -1 2 is not defined, since there is no number or angle whose sine is 2. If x is a number such that \x\ ^ 1, then y certainly exists. More- over, because of the periodicity of the sine function, there are infinitely many values of y = sin * x corresponding to every such value of x. For example, it x — 1/2, then y = sin -1 1/2 means that y is any number or angle such that sin y = 1/2. Then y may be taken as 7r/6, 57?/6, or any value that differs from these by integral multiples of 2tt. The totality of these values of sin -1 1/2 may be represented as — + 2mr and — + 2mr, where n — 0, ± 1, zb 2, =fc • • . We shall find it convenient to plot the graph of y — sin -1 x for a further study of the inverse function. Since y = sin 1 x and x = sin y express exactly the same relation between x and y, the graph of Fig. 8-2 may be used as a graph of the inverse function. We obtain the graph of y = sin - 1 x simply by interchanging the axes in Fig. 8-2 and reversing one of them in direction. The result is shown in Fig. 8-10. The question now arises whether the y-axis can be subdivided into intervals within each of which y has just one value correspond- ing to each x such that \x\ ^ 1. One way of doing this is by select- ing a first interval Other intervals, such as -- ^ y g — and — ^— £ y £ — =- > are then selected. With the entire 2/-axis thus subdivided, we may think of the graph of y = sin 1 x as consisting of the graphs of infinitely many single-valued functions or branches. 158 Trigonometric Functions; Inverse Functions IF Sec. 8-6 ^•■iSin~*a? +»X y** cos"" 1 x *»X Fig. 8-10. Fig. 8-11. To avoid any ambiguity in later applications as a result of this multiple-valued property of the inverse sine, we shall often restrict y so as to make the function single- valued. There will then be but one value of y corresponding to each value of x such that sin y — x. We shall determine this value from the branch for which — -- ^ y ^ — « This branch is called the principal branch. The values of y chosen from this branch are called the principal values of the inverse-sine function and are represented by the equation y = Sin -1 x or y = Arc sin x. Note that in this case the initial letter of the name is capitalized. This restriction to the two quadrants containing the smallest numerical values of y results in a single-valued function y = Sin 1 x. The values of y are such that - ■— ^ y ^ ~ f or x in the interval -1 ^ x ^ 1. For example, we have Sin- 1 (-1) = -tt/2, Sin- 1 (-1/2) = -tt/6, Sin- 1 = 0, and Sin" 1 1 = tt/2. The Inverse Cosine. We shall next consider the function y = cos x. With the help of Fig. 8-3, we see that the range ofy = cos x is the interval — 1 ^ y ^ 1, and that for every y in this interval there are infinitely many values of x such that y = cos x. We are thus led to the inverse cosine function, which makes correspond to y all the values of x such that y = cos #. If again we interchange the sym- bols x and y, we may write the inverse cosine function as r-l y = cos^ 1 x or = arc cos x. Sec. 8-6 Trigonometric Function^; Inverse Functions 159 This inverse function has the following properties. If # is a number such that \x\ > 1, then y does not exist, because there is no value of y for which |cos y\> 1. If x is a number such that \x\ ^ 1, then there are infinitely many- values of y designated by cos -1 x. The graph ofy = cos -1 x is shown in Fig. 8-11. The method for plotting it is similar to that used for graphing y = sin -1 x. We first write x — cos y. We then plot a cosine curve by proceeding as for Fig. 8-3, except that values of the independent variable y are laid off on the 7/-axis. The principal branch of the curve in Fig. 8-11 is the portion of the curve for which g y ^ 7r. It is represented by the principal value of the function, which is denoted by y — Cos" 1 x or y = Arc cos x. The Inverse Tangent. The inverse tangent function is y = tan" or y = arc tan x. H 1 H- -3 -2 -1 7T/2 -7T/2 y»tan _1 x H 1 h- O l 2 3 Fig. 8-12. It is represented by the graph in Fig. 8-12. We note that there are infinitely many valu.es of y for every value of x. The principal branch of the graph ofy = tan- 1 x is the portion 7T 7T for which — ~ <y <-~ * This is represented by the equation y = Tan -1 x or y = Arc tan x. 160 Trigonometric Functions; Inverse Functions Sec. 8-6 The Inverse Cotangent, Secant, and Cosecant. The other inverse trigonometric functions are : y = cot -1 x or y = arc cot x, y = sec -1 x or y = arc sec #, 2/ = esc -1 re or y = arc esc #. **x Fig. 8-13. We shall show only the graph of y = cot" 1 a?. See Fig. 8-13. The principal branch of this curve is given by < y < ir. Principal Values of the Inverse Cosecant and Secant. The selection of principal values of y = esc 1 x and y = sec 1 & is by no means uni- form among all authors. Some writers adopt the range between and 7r/2 for both functions when x is positive, and between —tt and — 7r/2 when x is negative. The authors prefer, however, to use the definitions Csc" 1 x and Sec -1 x = Cos -1 We have, therefore, the following ranges of principal values of the inverse trigonometric functions : -\ SSiirisSf. - -5- Si Csc-1 * si ^ Cos" 1 a: ^ 7r, < Cot" 1 x < ir, (Csc* 1 x * 0), ^ Sec" 1 x g tt (Sec" 1 z ^ y) y < Tan- 1 x < j > Sec, 8-6 Trigonometric Functions; Inverse Functions 161 Example 8-3. Find the value of cos (Sin -1 -= J • Solution: Let Sin" 1 -= = 0. Then sin 6 = ■= > and cos = ± ^ • Since only the 3 principal value of sin -1 r is used, 6 is a first-quadrant angle and cos 6 cannot equal — -= • Hence, cos I Sin" 1 - j = - • Example 8-4. Find the value of sin [Tan" 1 ( - as)], x being a positive number. Solution: Let Tan" 1 ( - x) = 0. Then tan = — a;, and lies between - x/2 and 0. If angle is constructed in standard position, as shown in Fig. 8-14, then sin is found to be — x — x Vl + x 2 Hence, sin [Tan -1 ( - x)] = Vl +z 2 **X Example 8-5. Find Arc cos (cot 60°). Solution: Let Arc cos (cot 60°) == 0. Then cos = cot 60° = .5774, by Table II. Therefore, the value of is Arc cos .5774. Since is restricted to the interval from 0° to 180°, must, in this case, be a first-quadrant angle. Hence, = Arc cos .5774 = 54°44', and Arc cos (cot 60°) = 54°44'. EXERCISE 8-2 In each of the problems from 1 to 6, find the inverse of the given function. l. y = 2x_-5 3 A 4 _i7 4 ^ = 15 X+ 6 2.2/ = 5. y = 2x - 1 6 2x - 1 Sx + 6 3. y = mx + b. 6. y = 2x - | • 3 4 In each of the problems from 7 to 22, find the value of the given expression. 7. Sin" 1 1 11. Cot" 1 0, 15. Tan 3 8. Tan-' 1. 12. Sin- (- 1). 16. Cos-- (^) 9. Arc cos V3 13. Cot -.VI. 3 10. Cos- 1 (- 1). 14. Csc-» 1. 19. Cos- 1 (-&■ 21. Sin" 1 ^-0.414). 17. Tan" 1 ( - y/3). 18. Arc tan (- 1). 20. Tan- 1 [sin (- ir/2)]. 22. Tan- 1 (-1.414). In each of the problems from 23 to 37, evaluate the given expression. 23. tan (sin" 1 j|) • 24. sin (sin- 1 1) • 25. sin (Tan-* ~) 26. cot (| Sin- 1 ^) 27. cos ((tar 1 1) • 281 tan (Sin-* .6450). 1 62 Trigonometric Functions; Inverse functions Sec. 8-6 29. sin (Cos- 1 .9200). 30. tan (Cot-* 2). 31. sec (sin- 1 i) • 32. sin (Cot-* i) • 33. cot (Cos" 1 1) • 34. sin (Cot~* |) • 35. cos [Cot-* (-|)] # 36. cos (sin- 1 |) • 37. Sin- 1 (tan ^) • In each of the problems from 38 to 56 simplify the given expression, taking u as a positive number. In 38 to 40, 54, 56, u < t rr/2. In 44, 46, 48, 55, u <1. 38. Cos- 1 (sin u). 39. Sin" 1 (- sin u). 40. Sin- 1 (cos u). 41. esc f Sin- 1 - j • 42. sin (Sin- 1 u). 43. tan (Tan" 1 v). 44. sec (Sin- 1 y/T^t?). 45. cot (Tan- 1 * — \ • 46. tan (Cot" 1 - = "j • 47. tan (Sin" 1 = \ • V \/l - ^ 2 ' ^ V^ 2 + 1/ 48. esc (Sin- 1 y/l - u*). 49. tan (Cos" 1 J: ") • 50. cos (Tan- 1 w). 51. sec (Cos -1 w). 52. cot (Sin- 1 u). 53. sec (Sin- 1 u). 54. Sin- 1 (sin u). 55. Cos" 1 (cos y/l - w 2 ). 56. Cot- 1 f tan y U "j • V Vi + w 2 / In each of the problems from 57 to 65, draw the graph of the given function by changing from the inverse function to the direct function and using the period, amplitude, and phase angle of the function to assist in plotting. 57. y = jr-sin-* x. 58. y = 2 tan -1 x. 3 1 59. y = - cos" 1 a;. 60. ?/ = x sin" 1 » — 1. 61. y = g cos- 1 x + 2. 62. y = 4 tan" 1 x 4- 3. 63. 2/ = — (Cos- 1 2x + 1). 64. y = 3 sin" 1 (2x + 1) - 2. 7T 65. 2/ = 4 tan- 1 (^-y^) 66. Prove that sin (Cos- 1 u) £ 0, if < w £ 1. 67. Prove that cos (Sin- 1 u) > £ 0, if < a ^ 1. 68. Prove that cos (Tan- 1 u) > 0, if u £ 0. 69. Prove that tan (Cos- 1 it) £ 0, if < u £ 1. 70. Prove that Sin- 1 u + Sin- 1 (- u) = 0, if - 1 £ t* £ 1. 71. Prove that Tan- 1 u -f Tan- 1 ( - w) = for all values of w. 72. Prove that Tan- 1 u + Tan" 1 (1/u) = y i if u > 0. 73. Prove that Cos" 1 w 4- Cos" 1 (-1*)=*-, if-l£u£l. 74. Prove that Tan- 1 u - Tan~* v = Tan- 1 , tt ,"" - > if u > and v > 0. 1 + uv V Linear Equations and Graphs 9-1. SOLUTIONS OF SIMULTANEOUS EQUATIONS Often problems arise that involve two or more unknowns and as many equations. The solution of such a problem requires the deter- mination of numbers which simultaneously satisfy the given equa- tions. Equations for which we seek common solutions are referred to as a system of simultaneous equations. If the system has at least one solution, it is said to be consistent; otherwise, it is called inconsistent. Let us investigate the following pair of simultaneous linear equations : f a\x + buj = ci, (9-1) { a 2 x + b 2 y = c 2 . It is desired to find all pairs of values of x and y which satisfy both equations, excluding from consideration the cases ai = &i = and On = b 2 = 0. We proceed by multiplying both sides of the first equation by b 2 and both sides of the second by b u obtaining afax + biboy = cib 2 , a 2 bix + b 2 biy = c 2 b\. Subtracting the second equation of this pair from the first, we obtain (9-2) («i?>2 — a 2 b\)x = cib 2 — C2&i- If (ai&2 — a 2 b\) t* 0, [we find that c\b 2 — c 2 bi ^ aib 2 — a 2 b\ Thus, we see that we have exactly one value for x in any solution which may exist. 163 1 64 Linear Equations and Graphs Sec. 9—1 By multiplying the original equations by a? and a u respectively, ^nd subtracting the first equation thus obtained from the second one, we obtain (9-3) (aib 2 — d2bi)y = a x C2 — «2^i. If (a x b 2 — ^261) ** 0, we find that a\C2 — CL2C1 y — • a x b 2 — «2&i Again, we have exactly one value for y in any solution which may exist. Hence, if (a x b 2 — &2&1) is not zero, we have at most one solution of the pair of given equations, namely, (9-4) *' x = Clh2 - C2bl 1 y = ai ° 2 ~~ a2Cl ■ d X b 2 — ^2^1 «1?>2 — #2^1 Substitution of these values for a; and y in (9-1) shows that we really have a solution. The reader should verify this solution by mak- ing the substitutions. Consistent and Independent Equations. If {a x b 2 — (hbi) ^ 0, we have exactly one solution of (9-1), which is expressed by (9-4), and the given equations are called consistent and independent We may consider, as an example, the pair of equations ( 2x + 3y = 24, 1 bx - 2y = 22. Here a x b 2 - (hb x = (2) (-2) - (5) (3) = -19, and the values x = 6 and y = 4 give a solution. In other words, when these values are substituted in the two given equations, both equations are satisfied. Thus, J 2(6) + 3(4) = 24, 1 5(6) - 2(4) = 22. Cases with a x b 2 — chb x = 0. Let us now consider cases in which a x b 2 — a%b x = 0. If a x ¥" 0, division by a x gives b 2 = — 01. a x We then define k = — > and we have (9-5) a 2 = kau fa = ft&i. Now let #i = 0. Then, since 61 cannot be zero by our initial assump- tion, and since a 2 b x = 0, we have a 2 = 0. In this case, if we define k = ~ 1 we see that (9-5) again follows. Thus, a x b 2 - a 2 b x = has 01 Sec. 9-1 Linear Equations and Graphs 1 65 been shown to mean that the left members of (9-1) are proportional. We note also that if a x b 2 — 0261 = 0, equations (9-2) and (9-3) become f • x = C1&2 — C2&1, (9-6) { • y = a x c 2 - a 2 ci. It is clear that these equations cannot be satisfied by any pair of values of x and y, unless both right members are also zero. Accord- ingly, if we wish to determine whether or not the given system has a solution when a x b 2 - a 2 b 1 - 0, it becomes necessary to take into account the two cases of the right sides of (9-6) being zero or not zero. Consistent and Dependent Equations. In the first case referred to in the preceding paragraph, a x b 2 — chbi = 0, and c 1 b 2 — c 2 b t and a x c 2 — a 2 Ci are also equal to zero. Hence, substituting the values of b 2 and a^ from (9-5), we have {c\k — 02)61 = (cifc — c 2 )a\ = 0. Since a u bi are not both zero, it follows that c 2 = kc x . This means that one of the original equations is a constant multiple of the other, and any pair of values of x and y that satisfy one equation will also satisfy the other. Under this condition, the equations are said to be consistent and dependent. We have as an illustrative example the equations J 2x + 3y = 24, [ 4z + 6?/ = 48. Here a x b 2 - a 2 b x = (2) (6) — (4) (3) = 0. Also, the coefficients of the unknowns and the constant term in the second equation are multiples of the coefficients of the unknowns and the constant term in the first, and the multiplier is 2 for all three terms. Infinitely many solutions exist. Some of them are : x = 0, y = 8 ; x = 12, y = ; 24 — 2t and x = t,y = — - — > where t is any number. d % Inconsistent Equations. Finally, let us suppose that in the orig- inal equations a x b^ — c^bi = and at least one of the numbers C\b 2 — c 2 bu a>ic 2 - CL2C1 is different from zero. Hence, no solution of (9-1) exists and the equations are inconsistent. This case is char- acterized by the condition that one of the numbers {c x k — (%) &i> (c x k — c 2 )di is not zero, in view of (9-5). It follows that c 2 ¥* kc x . 1 66 Linear Equations and Graphs Sec. 9-1 Hence, the multiplier for the left members of (9-1) does not apply po the constant terms. Consider, for example, the equations (2x + Zy = 24, [ 4x + &y = 7. Here the coefficients of the unknowns in the second equation are multiples of those in the first equation, and the multiplier is 2 for both terms; however, the multiplier for the constants is not the same as that for the other terms. Hence, these equations are inconsistent. 9-2. ALGEBRAIC SOLUTION OF LINEAR EQUATIONS IN TWO UNKNOWNS To solve a consistent and independent system of two linear equa- tions in two unknowns, we reduce the system to one equation in one unknown by eliminating one of the unknowns. The following example will illustrate two commonly used methods for eliminating the unknowns. Example 9-1. Solve the equations 2x+3y= 24, 5x - 2y = 22. Solution: Since aM - a 2 bi = (2) (- 2) - (5) (3) = - 19, the equations are consistent and independent, and there is but one solution. If we use the method of elimination by addition or subtraction, the procedure is the same as that indicated in obtaining the solution (9-4) from equations (9-1). To eliminate y, multiply the first equation by 2 and the second by 3, in order to make the coefficients of y numerically equal in both equations. We thus obtain 4a; + 6y = 48, 15a; - 6y = 66. Adding, we get 19a; = 114. Solving for x, we have x = 6. Now, substitute 6 for x in the first of the original equations. Then 32/ = 24 - 2a; = 24 - 12 = 12, or y =4. Alternate Solution: If we use the method of elimination by substitution, we begin by solving the first equation for y in terms of x. We thus get 24 -2a; Sec. 9-3 Linear Equations and Graphs 1 67 24 — 2x We then substitute — = for y in the second equation and obtain 3 Solving for x, we have or 5, - 2 (?!=*;) = 22. 15x - 2 (24 - 2x) = 66, 15x - 48 + Ax = 66. 19z = 114, Hence, and x = 6. Substituting 6 for #, as before, we find that y = 4. Example 9-2. A grocer has some coffee selling at 80 cents per pound and some at 90 cents per pound. How much of each must he use to get a mixture of 100 pounds worth 86 cents per pound? Solution: Let x = number of pounds of 80-cent coffee, and y = number of pounds of 90-cent coffee. Then * + y = 100, and 0.80a; + 0.902/ = 0.86 (100). Simplifying, we have x + y = 100, Sx + 9y = 860. These equations have the single solution x = 40, y = 60. 9-3. LINEAR EQUATIONS IN THREE UNKNOWNS In the solution of a system of three equations in three unknowns, one method is to employ the following steps, which we do not justify here : 1. Eliminate one of the unknowns from a pair of the equations; then eliminate this same unknown from another pair of the original equations. 2. Solve the resulting two equations for the two remaining unknowns. 3. Substitute the values found in step 2 in any one of the orig- inal equations to find the third unknown. Example 9-3. Solve the system of equations 2x + 2y - 2=5, x — by + 2z = 1, Zx + y — 4« = — 1. 168 Linear Equations and Graphs Sec. 9-3 Solution: Eliminate z from the first and second equations by addition to obtain 5s 4- y = 11. Now eliminate z from the second and third given equations by addition to obtain 5s — 9y = 1. We then consider the equations 5s + y = 11, hx — 9y = 1. Solving these equations for s and y by subtraction, we have x = 2 and 2/ = 1. Substitution of these values in the first given equation gives 2=2. Hence, the solution of the given system is x = 2, y = 1, z = 2. EXERCISE 9-1 In each of the problems from 1 to 30, solve the given system of equations. Check all solutions. L 4. 7. 10. 13. 16. 18. 20. 22. 24. 3s - 2y = 6, a; - 3y = 4. 3s + 2y = 1, x - 2y = 5. 3s + 2y = 4, 2s - 3y = 3. s + 2?/ + 1 = 0, x - 4y + 2 = 0. 11. 3s - y = 7, 2s 4- 2/ = 8. 2s 4- 3y 4- 1 = 0, 3s - 2/ + 7 = 0. x + 2y = 3, 2s 4- 3$/ = 1. 2s + Sy - 1 = 0, 3s 4- 2/ + 3 = 0. f 1 = 3, 2,37 3 , 1 _1 .-8 x+ 4 y= 9 - [2 X+ S V """2 3s - 2/ + 2« = 4, * + 2y - 3* = 1, 2s - Sy + 2=2. s + 2y + 32 + 1 = 0, x - 42/ + 52 4- 1 = 0, 2s + 62/ + 72 + 2 = 0. s 4- 22/ - 32 + 1 = 0, - x 4- 32/ - 42 - 5 = 0, 2s 4- 62/ - 42 4- 3 = 0. 3s + 2y + z =2, 5s + 2/ 4-3=0, 2x - 3?/ - 42 4- 5 = 0. 3s - y 4- 42 = 2, 4s 4- 4y 4- 42 = 5, >2s - y 4- 62 = 9. 17. 19. 21. 23. 25. 3. \y - 3s = 6, }s 4- 22/ = 3. 6. (2s 4- 2y - 3 = 0, \ 5s 4- 32/ 4- 4 = 0. 9. |3s + 2/ + 7 = 0, [4s 4- 82/ 4- 9 = 0. 12. f 2s 4- 62/ - 7 = 0, J3s - 82/ 4- 9 = 0. 15. f 2x 4- 42/ - 52 = 3, ' 3s 4- y - 72 = 2, ,4s 4- Sy - IO2 = 1. s 4- 22/ - 2=3, - s 4- 42/ 4- 22 = 1, 3s 4- 2/ - 32 = 2. - 3s 4- 82/ 4- 92 - 3 = 0, 2s 4- 32/ 4- 2-4=0, 3s - 22/ - 22 - 4 = 0. 3s - y 4- 52 = 0, s - 42 = 2, 4s - 22/ - 32 4- 1 = 0. 3s - 42/ 4- 22 = 3, 2s 4- y =1, 5s - 3i/ 4- 42 4- 5 = 0. (2s 4- Sy - 52 4- 2 = 0, 52/ 4- 32 - 2 = 0, 1 3 n *- 2/-2*~2 =0 ' Sec. 9-4 Linear Equations and Graphs 169 26. 3y + 2z = 4, 2x - 2y + 3z = 3, I3x + 4a = 2. 27. 28. ,2 2 1 x y z x y z ' = 1. .-2+1=1 x 2/ 3 29. 30. ±+2+1 I3(x +2/) -4(s 1 4(3 + 2/) - 3(3 3(3 + y) - 4(3 4(3 - 2/) — 3(3 For each of the following systems of equations, determine whether it is and independent, consistent and dependent, or inconsistent. 31. 34. 37. 40. 2x - 3y = - 5, 32. 43 - 6y = -- 3. 2z + 4y = 3, 35. x + 22/ = 6. 2z - 3?/ = 1, 38. 4z + 6?/ = 2. - 3 + 3z/ = 2, 41. 23 = 62/ 4- 14. (33 - 5y + 8=0, \ x + 82/ - 10 = 0. ( 23 - 72/ + 1 = 0, \2ly -63 -3 =0. [ - 93 + 122/ = 3, 33 + 42/ = 1. - 2/) = 5, - y) = 5. - y) = 5, + 2/) = 7. consistent = 8, + r=C lay + 63 = a&. 63 + ay = a6c. 9-4. GRAPHS OF LINEAR FUNCTIONS The discussion of rectangular coordinates in Section 2-1 set the stage for the pictorial representation of a function. By this repre- sentation of a function f(x), we mean the graph of the equation y = f(x). It consists of all points, and only those points, whose coordinates x and y satisfy the equation. In the same section we considered the graphing of lines parallel to the coordinate axes. The equation x = 3 was shown to represent a vertical line, that is, a line parallel to the y-axis which intersects the #-axis at the point (3, 0). This line thus includes all points 3 units to the right of the y-axis. This example illustrates the fact that a linear equation in x alone represents a line that is parallel to the 2/-axis. Similarly, y = 2 was shown to be the equation of thfc horizontal line whiph is parallel to and 2 units above the #-axis. And this example illustrates the fact that an equation in y alone represents a line parallel to the z-axis. Furthermore, it is proved in analytic geometry that the graph of every first-degree, or linear, equation in x and y is a straight line; and, conversely, that every straight line is the crraDh of a linear equation. 170 Linear Equations and Graphs Sec. 9-4 We shall proceed by first preparing a table of corresponding values in a given problem and then plotting the corresponding 'points on the coordinate system to obtain the graph of the equation. The following illustrative examples will point the way toward an understanding of the procedure in the graphing of linear equations. Example 9-4. Graph the function 2x + 3. Solution: Let y = 2x + 3. Then assign any values for x ) substitute them in the equation, and obtain the corresponding values for y. The table and the graph are shown in Fig. 9-1. Since a straight line is definitely determined when two points are known, only two pairs of values of x and y are needed in graphing a linear equation. We can, how- ever, use three points in order to check our work. X y 3 1 5 -2 -1 *~x Fig. 9-1. Fig. 9-2. Example 9-5. Graph the equation 2x + 3y = 6. Solution: The equation may be solved for y in terms of x. Then y = -|a? +2. A table of corresponding values of x and y and the graph are shown in Fig. 9-2. 9-5. INTERCEPTS In general, the points where a curve crosses the coordinate axes are the easiest to obtain. The ^-intercepts are the values of x at the points where the graph crosses the #-axis. Since y = on this axis, the ^-intercepts are the values of x that correspond to y = 0. Similarly, the ^/-intercepts are the values of y at which the graph crosses the y-axis. They are the values of y that correspond to x = 0. Hence, we have the following rule : To find the ^-intercepts, set y = in the equation and solve for x. To find the ^/-intercepts, set x = in the equation and solve for y. Sec. 9-6 Linear Equations and Graphs 171 Example 9-6. Find the intercepts of the line 2x + 3y = 6. Solution: To find the x-intercept, let y = 0. Then 2x = 6, and x = 3. To find the ^/-intercept, let x = 0. Then 3y = 6, and 2/ = 2. Note that the intercepts a; = 3 and y — 2 found in this solution correspond to the points (3, 0) and (0, 2), respectively, where the line in Fig. 9-2 crosses the coordinate axes. 9-6. GRAPHICAL SOLUTION OF LINEAR EQUATIONS IN TWO UNKNOWNS In the graphical solution of two linear equations in two unknowns, the graphs of the two equations are drawn with reference to the same coordinate axes. Since the solution of two equations in x and y is a pair of values of x and y which satisfy both equations, the solution must represent graphically a point common to both lines represented by the equations. Hence, the values of x and y which satisfy both equations give the coordinates of the point of inter- section of the lines. We find, therefore, that the two lines intersect in a single point, are parallel, or are identical, according as the equations are consistent and independent, inconsistent, or consistent and dependent. Example 9-7. Solve graphically 2x + 3y = 24, 5x - 2y = 22. Solution: Tables of corresponding values for the two equations and also their graphs are shown in Fig. 9-3. It is seen from the graphs that the lines intersect at the point (6, 4). That x = 6, y = 4 gives the solution of the given equations may be checked by substitution. Y 2x + dy = 24 bx - 2y = 22 12 x y -li 22 5 **x Fig. 9-3. 172 Linear Equations and Graphs EXERCISE 9-2 Sec. 9-6 In each of problems from 1 to 9, graph the given function. In each case give the a:- and ^-intercepts. 1. 4a: - dy = 1. 2. ?/ = 2s - 8. 3. y — x - 4. 4. y = x + 5. 5. 2/ = 3a\ 6. 2/ = z. 7. y = 3a: -f 4. 8. y = a; - 3. 9. y = 3a; — 5. Solve each of the following systems of equations graphically. 10. f x - 3y = 1, 11. 132/ -2a; =0, 12. r 2?/ = x -3, [3a: - 2y = 0. \ x + y = 2. [2a: = ?/ + 3. 13. (2?/ = a; + 3, 12a: = 16 - 3?/. 14. (3a; - y = 4, I 2/ - 3x = 1. 15. [ - =8, x -y 3 -4 l2(x - y) 16. (2a: - 0.-4, 13a; + 2?/ = 12. 17. !2x +4 = 5# - 3, 13?/ -4 =4a; +2. 18. .2a: - 3y = 12. 10 Determinants 10-1. DETERMINANTS OF THE SECOND ORDER Let us consider the following system of two linear equations in two unknowns: , . f aix + b\y = ci, (10-1) The solution of these equations by the method of Section 9-1 is given by (10 2) x = — r r- > 2/ = — i — " i— * It is understood that t^fto — ffa&i "^ 0. We shall at this point introduce a more convenient way of writ- ing the expression iiib 2 — o-j&i. The notation which we select for this purpose will enable us to express also the numerators of (10-2) by means of the same symbol with the proper changes in letters. In choosing a symbol we shall select a form which will exhibit the numbers a u b u a*, b> 2 in the same relative positions as in (10-1) . Thus, we write Go 62. This arrangement of the four numbers in a square array, consist- ing of two rows and two columns, is then enclosed within vertical bars, as follows : oi 61 a 2 bo This symbol represents a determinant of second order. Thus, we start with a square array, or matrix, such as \a 2 b 2 / 173 174 Determinants Sec. 10-1 We then associate with this array a number ai& 2 — & 2 &i, called its determinant, which is denoted in the following manner : The numbers a u b lf a 2 , b 2 are called elements of the determinant. The numbers ai and b 2 lie on the principal diagonal; the numbers a 2 and 61 lie on the secondary diagonal Note. We observe that the "expanded" value of the foregoing determinant is equal to the product of the elements on the principal diagonal minus the product of the elements on the secondary diagonal. It is interesting to note also that this value is the alge- braic sum of all possible products obtainable by taking one and only one element from each row and one element from each column. Each product is preceded by a plus sign or a minus sign, according to a rule to be stated in Section 10-2. Using the notation of determinants, we can write the solution (10-2) of (10-1) in the form (10-3) C\ h C2 62 O-l 61 0.2 b 2 ai Cl G2 C2 ai 6l d2 62 We note that the value of each of the unknowns in (10-3) may be written as a fraction whose denominator is the determinant of the coefficients as they stand in (10-1), and whose numerator is the determinant formed from that of the denominator by replacing the column of coefficients of the unknown in question by the column of constant terms. Note. If a 2 = ka x and b 2 = kb lt where k is any number, then 61 a\ a 2 62 = 0. In this case, the equations of the system (10-1) are inconsistent unless both numerators of the fractions in (10-2) are also equal to zero, that is, unless c\ 61 C2 &2 = and ai Ci «2 C\ = 0. Therefore, the equations of the system (10-1) represent distinct, parallel straight lines if c 2 ¥= kc u or they represent coincident lines if Co = kcu Sec. 10-2 Determinants 175 Example 10-1. Solve the system of equations 4x + Sy = 1, Sx - 2?/ = 22. Solution: Using determinants, we have -2-66 -8-9 Also, 2/ = 1 3 22 -2 4 3 3 -2 4 1 3 22 4 3 3 -2 68 = 4. 17 88 85 = -5. - 8 - 9 - 17 10-2. DETERMINANTS OF THE THIRD ORDER A determinant of the third order is a number designated by a square array of nine elements arranged in three rows and three columns and enclosed within vertical bars. An example is a\ bi c\ (10-4) D = 0,2 62 C2 «3 b-s ra The value, or expansion, of the determinant (10-4) is defined as the quantity (10-5) D =ai &2 C2 -6l a 2 c 2 + Ci a 2 62 63 C3 «3 C3 #3 63 or as the quantity (10 0) D = tti62C3 — 01&3C2 + ^1^203 — &1C52C3 + Clfitefo — Ci&2«3, Here the products such as aib 2 Cs 9 aib 3 c 2 , and biases are known as the terms of the determinant. Minors and Cofactors. In any determinant the minor of a given element is the determinant of the array which remains after delet- ing all the elements that lie in the same row and in the same column as the given element. Thus, in (10-4) the minors of a u 61, Ci are, respectively, 62 C2 0,2 C2 CL2 62 63 £3 as C3 as 63 176 Determinants Sec. 10-2 The cof actor of an element which lies in the ith row and ftth column is equal to the minor of that element if i -f k is even, and is equal to the negative of the minor if i + k is odd. That is, cof actor = (— l) i+k • minor. Thus, in the determinant in (10-4), the cof actor of a x equals the minor of ai, since a x lies in the first row and in the first column and i-ffc = l + l = 2, which is even. Similarly, the cof actor of 6i is the negative of the minor of b lf since i + fc = l + 2 = 3, which is odd. Often the following procedure may prove more convenient for finding the sign corresponding to a given element. Beginning with + in the upper left-hand corner, change sign from place to place, moving horizontally or vertically, until the position for the element in question is reached. The schematic arrangement of signs cor- responding to the elements of a third-order determinant is thus as follows : + — + — + — + — + Note that the sign for any position is independent of the path followed in arriving at that position. We shall designate the value of the cof actor of an element by the corresponding capital letter, and we shall use the subscript that occurs with the element itself. Thus, the cofactors of a u b u c x are, respectively, (10-7) Ai = b 2 bs C2 C'S Bi - - «2 C2 , Ci = «2 &2 a,3 C's «3 b' ti Hence, (10-5) for the expansion of the determinant may also be written as follows : (10-8) D = aUi + biBi + c x Ci. This sum is called the expansion of the determinant according to the elements of the first row. We observe at this point that the right member of (10-6) repre- sents all possible products, here 3 ! in number, that can be formed from the determinant in (10-4) by taking one and only one element from each row and each column. It follows also that the value of the determinant is the same, regardless of the row or column Sec. 10-3 Determinants \77 according to which the expansion is made. Thus, we may express the determinant as (10-9) D = biBi + b 2 B 2 + &3#3, or as (10-10) D = a 3 A 3 + 63B3 + C3C3. These equations represent the expansion according to the elements of the second column and according to the elements of the third row, respectively. Example 10-2. Expand the determinant 2-5 3 6 - 1 2 1 7 4 according to the elements of the first row and according to the elements of the third column. Solution: The expansion according to the elements of the first row is 2 1 + 5 6 1 + 3 6 2 7 4 - 1 4 - 1 7 This reduces to 2(8 - 7) + 5(24 + 1) + 3(42 + 2) = 259. Expanding according to the elements of the third column, we have = 3(42 + 2) - (14 - 5) + 4(4 + 30) = 259. (5 2 -1 7 -1- 2 -5 -1 7 + 4 2 -5 6 2 10-3. PROPERTIES OF DETERMINANTS From the definition of the value of a determinant we may deduce the following important properties of determinants. These proper- ties supply us with more convenient methods for evaluating a determinant. Note. For a more complete discussion of these properties, the stu- dent is referred to any one of the various treatises on determinants or to texts on the theory of equations or on solid analytic geometry, where he will also find proofs which apply to determinants of any order. The properties listed here will be employed in examples that fol- low, and their usefulness in simplifying determinants will be illustrated. 178 Determinants Sec. 10-3 Property 1. The value of a determinant is not changed if its rows and columns are interchanged. Property 2. If all the elements of a row, or of a column, are multi- plied by the same number, the value of the determinant is multiplied by that number. For example, ka\ b\ = fc a x 61 ka,2 62 0,2 62 Property 3. If two rows, or two columns, of a determinant are identical or proportional, the value of the determinant is zero. For example, let the first two columns be identical, as in the determinant a\ ai c\ D = 0,2 CL2 C2 03 03 c$ Then, expanding according to the elements of the third column, we have D = a 02 02 - C 2 01 Ol + C 3 01 Ol O3 03 03 03 02 02 = 0. Oi 61 Ci a 2 b 2 c 2 - 0.3 63 C3 Property 4. The value of a determinant is not changed if we add to the elements of any column (row) any arbitrary multiple of the elements of any other given column (row) . For example, rti + nb\ b\ C\ «2 + Tib2 62 C2 • «3 + W&3 b 3 C 3 The proof follows. Expanding according to the elements of the first column, we find that 01 + nbi 61 c\ tt2 + W&2 &2 C2 03 + nbz 63 C3 The last determinant vanishes, since two columns are identical. Ol 61 C\ 02 b 2 C2 + n 03 63 C3 61 61 C\ b 2 62 C2 h h C3 Sec. 10-3 Determinants 179 Example 10-3. Evaluate the determinant 4 3-1 D = 5 1 2 4 Solution: By adding 2 times the elements of the first row to the elements of the second row, we obtain 4 3 - 1 4 3 - 1 5 1 2 = 13 7 2 4 3 2 4 3 If now we add 3 times the elements of the first row to the third row, the determinant becomes 4 3-1 13 7 14 13 Expanding according to the elements of the third column, we have 13 7 (-D 14 13 = -71. In this example, we first converted the given determinant to one in which all but one of the elements of the third column are zero. For the final expansion, the given determinant was thus reduced to a determinant of the second order, and its value was easily found. Example 10-4. Without expanding the determinant, show that x = — 2 satisfies the equation 3x 2 x z - x 3 1 7 6-4 1 = 0. Solution: Substituting — 2 for x in the determinant, we have 12 -8 2 3 1 7 6 -4 1 This equals zero, since the first and third rows are proportional. Hence, the equation is satisfied by x = — 2. 180 Determinants EXERCISE 10-1 Sec. 10-3 In each of the problems from 1 to 12, evaluate the given determinant. 2 -5 3 -5 3-5 6-10 2 - 1 3 2 7 3 -5 3, 3 4-5 6. 10. 3 8 7. 6 -6 X y 1 Xi Vi 1 • 11. x 2 2/2 1 7 2 14 4 1 -3 2 2 1 4 -2 1-3 2 4-5 -6-1 4 4 8-9 12. 9 7 8 14 2 2 3-2-1 2 1 4 2 2 6 1 -6 3 5 7 15 ♦ In each of the problems from 13 to 19, solve the given system of simultaneous equations by means of determinants. Cheek all solutions by substitution in the equations. 13. f x + 3y = 5, \2x - 4?/ = 7. 14. \2x - y = 1, Ux +2y =4. 15. Sx + 2y = 5, 2x - Sy = 5. 16. If D represents the determinant in Problem 10, show that D = is the equation of the straight line through the points (x lt ?/,) and (x 2 , 2/2). 17. Find the ^-intercept and the ^-intercept of the line whose equation is x y 1 3 4 1 =0. 2-3 1 18. Solve graphically the following system of equations: = 0. 19. Find the coordinates of the vertices of the triangle whose sides are the straight lines 3-0+2=0, 2x + 9y + 15 = 0, and 7x + Ay - 30 = 0. 20. Find the vertices of the parallelogram formed by the following lines: = 0. X y 1 2 1 = 0, 1/2 1 1 x y 1 1 1 1/2 1 x y 1 2 1 1 = 0, -3 2 1 X y 1 x y 1 6 - 1 1 = 0, 1 3 1 = 0, 1 1 2 4 1 X y 1 5 1 1 1 -3 1 Sec. 10-4 Determinants 181 10-4. SOLUTION OF THREE SIMULTANEOUS LINEAR EQUATIONS IN THREE UNKNOWNS Let us consider the following system of linear equations : a x x + hy + az = d\ } (10-11) a 2 x + b 2 y + c 2 z = d 2 , , d3X + foy + c 3 z = dz. The determinant of the coefficients of the unknowns is a\ b\ c\ D = a 2 c 2 (iz bs C3 For the solution of such a system, we employ the following theorem, which is known is Cramer's Rule. Theorem. If the determinant D of the coefficients of the system is not equal to zero, the system has just one solution. In this solu- tion, the value of any unknown is equal to a fraction whose denom- inator is D and whose numerator is obtained from D by replacing the column of coefficients of the unknown in question by the column of constants d u d 2 , and d> 6 . Proof. Let the numerators of the fractions for x, y, z be denoted by D u D 2 , D 3 , respectively. We proceed to show that if equations (10-11) are to be satisfied, then (10-12) Dx = D u Dy = D 2> Dz = D 3 . Specifically, the equation for x will have the form ai b\ c\ (10-13) a 2 as b 2 c 2 C'S di bi ct d 2 62 C2 d 3 63 Cz Similar equations may be written for y and z. To find x, the first equation of the system in (10-11) is multi- plied by the cof actor A u the second by A 2 , and the third by A 3 . After adding and collecting terms, we obtain (10-14) (aiAi + a 2 A 2 + a%A$)x + (biAi + b 2 A 2 + foAs)y + (ciAi + c 2 A 2 + c$Az)z = d\A\ + d 2 A 2 + d^As. The coefficient of x in (10-14) is the expanded value of D accord- ing to the elements of the first column. The coefficient of y is biAi + b 2 A 2 + 63A3; 182 Determinants Sec. 10-4 this is equal to 61 h C\ 62 b 2 C2 &3 h C3 which equals zero, since two columns are identical. Similarly, the coefficient of z is zero. Hence, we have shown that the left side of (10-14) is the expanded form of the left side of (10-13) and that the two are therefore the same. The right side of (10-14) is cM-i + d 2 A 2 + d 3 A 3 , which is the expansion of the determinant d\ 61 c\ d>2 &2 C2 dz b 3 c-3 This determinant may be obtained from D by replacing the coeffi- cients of x by the column of constants ; that is, it is the expansion of the determinant that we have called Di. Hence, the equation Dx = D x in (10-12) is established. By a similar procedure we can show that the equations Dy = D 2 and Dz = D 6 are also valid. If D ¥* 0, the value of x is given by the equation (10-15) di h Cl d 2 62 C2 d 3 bs C3 X = tti 61 Ci a2 62 C2 tt3 63 C3 Similar values may be found for y and z. The proof is complete. The proof of (10-12) is valid whether D ^ or D = 0. If Z) ^ 0, the equations of the system (10-11) are consistent and have only one solution, which is of the form (10-15) ; that is, (10-16) y = D2 D z = D3 D ' If D = 0, and any one or more of the other determinants, D u D 2 , Z) 3 , is not zero, the given system of equations has no solution and is inconsistent. Sec- 10-5 Determinants 183 If D = 0, and all the other determinants are zero, the equations of the system (10-11) may be consistent or inconsistent. If they are consistent, there are infinitely many solutions. This case will be treated in Section 10-5. The following example will illustrate the case for which D ¥=0. Example 10-5. Solve the system of equations x - y + z = 1, x + y - 2z = 3, 2x - y -f dz = 4. Solution: Here 1-11 1-11 D = 1 1 -2=5; Di = 3 1 - 2 = 11; D 2 = 1 3 -2 =8: D s = 1 1 3=2. 1 -1 1 1 - 1 1 1 1 -2 = 5; Di = 3 . -2 2 - 1 3 4 - 1 3 1 1 1 1 - I 1 1 3 -2 = 8; Z>3 = 1 L 3 2 4 3 2 — L 4 D 8 Z) 2 Hence, x=^=~>y=z-j£~-iz=-^=-' If we check by substitution, we 11 5 D D find that these values satisfy the given equations. 10-5. SYSTEMS OF THREE LINEAR EQUATIONS IN THREE UNKNOWNS WHEN D = We note that when D = 0, the system (10-11) will not have a solution if any one of the other determinants D u D 2 , D 3 is different from zero. Suppose that a solution is given by x = r, y = 8, z = t. Then the equations (10-12) become r • = Z>i, 8 • = D 2 , * • = D 3 . It follows that Z?i = 0, D 2 = 0, Z) 3 = 0. The following example will illustrate the case of consistent equa- tions where D = and D x = D 2 = D s = 0. The equations are said to be dependent, and they have infinitely many solutions. The student should construct an example to show that the equations (10-11) may be inconsistent when D = 0, even though D x = D 2 = D 3 = 0. 184 Determinants Sec. 10-5 Example 10-6. Solve the system of equations x + y - z =3, 2x - 2/ + 3z = 1, 4z - 2y + 6z = 2. Solution: Here Z) = 1 1-1 2-1 3 4-2 6 This equals zero because the second and third rows are proportional. But, we also find that 1 D x = 3 1-1 1-1 3 2-2 6 = 0, D 2 = 13-1 2 1 3 4 2 6 = 0, 7) 8 = 1 3 - 1 1 -2 2 = 0. In this case the given equations have a solution. In fact, the second and third equations are proportional, and so either of these can be solved together with the first equation for two of the unknowns in terms of the third. For example, x = • 4 - 2z y =■ 5 +5z 3 ' y ~ 3 Thus, we have a single value of x and a single value of y for every value of z. However, there are infinitely many values of z, and therefore infinitely many solutions of the given equations exist. 10-6. HOMOGENEOUS EQUATIONS The system (10-11) is homogeneous if dx = 0, d 2 — 0, and d 8 = 0. Such a system always has the trivial solution x = y — z = 0. When d x = d 2 = rf 3 = 0, it is seen that J5 X = D 2 = Z) 3 = 0, for each of these determinants has zero for every element in one column. If D ¥= 0, it follows from (10-16) that we can have but one solution, which is given by Di _ n .. _ D 2 _ n . _ Z) 8 ^ = ^ = 0, » = -5= = ' = D 0. Hence, if the given system is to have a solution besides the trivial solution, D must equal zero. It may be shown that, if D = 0, non- trivial solutions always exist. Example 10-7. Solve the system of equations x - y + z = 0, 2z - 3y + 42! = 0, 5x - 2y - z = 0. See. 10-7 Determinants 185 Solution: The determinant D is 1 - 1 1 2 -3 4 = 5 -2 - 1 0-1 -1-3 1 3 - 2 - 3 = 0. Therefore, nontrivial solutions exist. To find these solutions, we proceed as follows: Transpose z in each of the first two equations, and solve for x and y in terms of z. Then we have D = Hence, 1 2 - 1 -3 = -1, Dt = Di -z - 1 -4z -3 = 2, and = - 2, Z> 2 = 1 2 — z -42 = -22. Substitution shows that these values also satisfy the third equation. The given system therefore has infinitely many solutions, and the values of x, y, and z are re- lated by the equations x = z and y = 2z. EXERCISE 10-2 In each of the problems from 1 to 6, solve the given system of simultaneous equations by means of determinants. Check all solutions by substitution in the equations. 1. {Sx + y - z = 11, 2. ( x- y-2z=-l, hx -2y = 0, 12s - 4?/ -f z=3. 2x - ?y - 3z = 7, 6. x 4- 2y - z = 10, 3s - 3*/ + 22 = - 7. For each of the following systems find at least one nontrivial solution, or show that there is no nontrivial solution. \3x + y - 2 = 11, 2. z + 3?/ - 2 = 13, [ a? + y - 3z = 11. \2x + y +32 = - 10, 5. 2s - 2// + 2=2, [6x + 2// -22 =5. a? - 2/ + 62 = 7, \2x + 3y + 62 = 0, ( £ + 2y + 92 = 3. [2j - 2/ = 3, 2x -32 = - 1, (32 - y = 2. 7. I 3 - 2i/ + 22 = 0, J 2s - by +2=0, [4z — 11?/ — 2=0. 8. 2s - 3?/ + 42 = 0, x +3y - z = 0, 7s + 3y + 52 = 0. 9. z - 2y + 32 = 0, 2x - ?/ + 42 = 0, 3.r + 2/ - 2=0. 10-7. SUM AND PRODUCT OF DETERMINANTS .Closely related to Property 4 of Section 10-3 is a theorem con- cerning the sum of determinants. We shall illustrate the theorem for splitting the elements of a given column into two parts by means of the following equality for third-order determinants : ai + 61 ci ^ Q>2 ~f" &2 C2 d,2 Q>3 + &3 C3 (#3 ci\ C\ d\ Q>2 C2 G?2 + 03 C3 dz 61 Ci rfi 62 C2 C?2 63 C3 C?3 186 Determinants Sec. 10-7 This can be shown to be true by expanding the three determi- nants according to the elements in the first column and noting that in the expansion the minors are the same for all three determinants. Example 10-8. Show that 2 4 5 1 4 5 3 4 5 1 - 1 + 4 - 1 = 5 - 1 3 7 6 - 1 7 6 2 7 6 Solution: Expanding each of the determinants on the left side of the equation according to the elements in the first column, we have - 1 7 6 - 4 5 7 6 + 3 _ 4 5 1 + - 1 7 G -4 4 7 5 6 - 4 5 - 1 = 3 - 1 7 G -5 4 7 5 6 + 2 4 - 1 5 • ai bi C\ d\ = + a 2 b 2 0,2 &2 This result is the expansion of the determinant on the right in the given equation according to the elements of the first column. Computing the value of each determi- nant in the given equation, we see that each side reduces to 47. A similar theorem for splitting the elements of a given row is illustrated by the following example : ai + Ci 61 + di CL2 &2 This can be shown to be valid by expanding according to the elements of the first row. Thus, consider the determinant 4 5 1 7 This may be written as a sum in various ways. Examples arc 2 3 + 2 2 and 4 5 + 4 5 1 7 1 7 -2 2 3 5 We shall state without proof the rule for the product of two determinants of the same order. The product is equal to a determinant of like order in which the element of the ith row and kth. column is the sum of the products of the elements of the ith row of the first determinant and the corresponding elements of the kth column of the second determinant. For example, for second-order determinants, we may write 1-5 +2-7 1-6 +2-8 1 2 3 4 5 6 7 8 3-5 + 4*7 3«6+4'8 Sec. 10-7 Determinants 187 That the product of the values of the two determinants on the left equals the value of the determinant on the right is checked by expanding. Thus, the desired equality becomes (4 - 6) • (40 - 42) = (950 - 946), or (-2). (-2) =4. To illustrate the multiplication of two third-order determinants, we have that 1 7 6 - 3 9 3 5-42 3 2 5 1 - 1 2 • 4 6 is equal to 3 • 1 + 2( - 3) + 5 • 5 3-6+ 2-3+5-2 3-7+2-9+ 5( - 4) 1- 1 +(- 1) (-3) +2-5 1-7 + (-1)9+ 2(- 4) 1 • 6 + (- 1)3 + 2 • 2 4-l+6(-3)+0-5 4-7+6-9+0(-4) 4-6+6-3+0-2 This reduces to 22 19 34 14 - 10 7 14 82 42 which is equal to — 630. Also, computing the values of the given factors, we have 30(- 21), which equals - 630. EXERCISE 10-3 In each of the first three problems, combine the given determinants into a single determinant, and evaluate the result. 1. 2 2 2 2 - 7 -3 - 1 7 + -3 - 1 7 • 4 8 - 10 4 8 - 10 1 2 2 1 4 2 2-3 1 - 2 2 1 • 4 8 - 1 4 19 - - 1 2 4 -5 2 4 -5 2 4 -5 -6 - 1 4 + - 6 - 1 4 + - 6 - 1 4 1 2 12 4 j 5 -6 - 1 3 - 16 188 Determinants Sec. 10-7 In each of the following problems, find the product of the determinants without evaluating the individual factors. 3 1 3 1 4. . 1 -2 1 5 11 21 12 3 6. 1 • ] . 2 3| 1 3 1 /I 12 3 -1|\ I \| — 1 4 5 l|/ I - (V - 2 . 4 %( \| 9 3 6 7|/ \ 2 6 3 8 13 21 5 4 2 - 1 3 1 -3 2 1 4 I)' -2 4 4 10 11 Complex Numbers 11-1. THE COMPLEX NUMBER SYSTEM There are many problems that cannot be solved by the use of real numbers alone. We observe, for example, that the equation x 1 + 1 = has no real root, since x 1 can never be negative if # is a real number. In order to provide solutions to such equations, a new system of numbers, called the complex number system, was intro- duced. Later in this book, we shall find many instances of solutions involving complex numbers. We shall now define a complex number as an ordered pair of real numbers, which we denote by (a,b). If the numbers a and b are regarded as the Cartesian coordinates of a point in a rectangular coordinate system, we have a one-to-one correspondence between the set of complex numbers and the set of points in a plane. The plane is called the complex plane. Two complex numbers (a, b) and (c, d) are equal if and only if they correspond to the same point, that is, if and only if a = c and b = d. Addition, subtraction, and multiplication of complex numbers are defined as follows : (11-1) (a, b) + (c, d) = (a + c, b + d), (11-2) (a, b) - (c, d) = (a - c, 6 - d), (11-3) (a, b) • (c, d) = (ac - 6d, ad + 6c). For example, (1, 3) + (5, 2) - (G, 5), (1, 3) - (5, 2) = (- 4, 1), (1,3). (5,2) = (-1,17). We also define the following special complex numbers: (11-4) = (0,0), (11-5) 1 = (1, 0), (11-6) % = (0, 1). 189 1 90 Complex Numbers Sec. 11-1 The complex number serves as a zero of the complex number system, while 1 serves as a unit, in accordance with the following properties : + (a, b) = (a, b) + = (a, b), • (a, 6) = (a, 6) • = 0, 1 • (a, 6) = (a, 6) • 1 = (a, 6). The so-called imaginary unit i= (0, 1) will be discussed in more detail in Section 11-2. If A; is a real number, we define (11-7) * • (a, b) = (fc, 0) • (a, 6) = (fca, kb). Also, we define (11-8) - (a, 6) = (- 1) • (a, b) = (- a, - 6). Since (a, 6) + (—a, —6) = (0, 0), the complex number —(a, b) is called the negative of (a, 6). The Reciprocal of (a,b). If (a, b) ¥= 0, then it has a reciprocal (#, #) such that (a, 6) • (x, ?/) = 1. Furthermore, the reciprocal is given by + 6 2 ' a 2 + b 2 ) By (11-3) and (11-5), the equation (a,, b) *\x,y) =1 may be written in the form (ax — by, ay + bx) = (1, 0). Since a 2 + b 2 ¥= 0, we may determine x and 2/ by solving the simul- taneous equations [ ax — by = 1, ( for + a?/ = 0. The values of x and 7/ can be foundry any of the methods taken up in Section 9-1. The solution is _ a __ b X ~ a 2 + b 2 ' y ~ ~ a 2 + b*l We have, therefore, verified (11-9) . Division of (a,b) by (c, d). If (c, d) ¥= 0, division can be defined as follows : (a, b) -v- (c, d) = (a, 6) • (u, «), where (w, t;) is the reciprocal of (c, d). By (11-9), we have (11-10) (a, b) + (c, d) = (a, 6) (^-^ , - ^5) (11-9) (X|V) =(«_,_ _^_) + cP c 2 + — ad -\- b c 2 + d 2 ' c 2 + d 2 _ /ac + bd — ad + &c\ ~ \c 2 + d 2 ' c 2 + d 2 ) Sec. 1 1 -2 Complex Numbers 1 9 1 Forexamp.e, (5, 13) * (3, - 2) = (- » , {§) . This result can also be verified by the method in Section 11-3. 11-2. THE STANDARD NOTATION FOR COMPLEX NUMBERS The special number i = (0, 1), defined by (11-6), has the follow- ing property : (11-11) t* = (0, 1) • (0, 1) = (- 1,0) = - (1, 0) = - 1. We shall now show that (a, 6) and the binomial form a + bi are equivalent, or that (11-12) (a, b) = a + bi, in which a + bi means al 4- bi. By (11-5), (11-6), and (11-7), a\ + bi = a(l, 0) + 6(0, 1) = (a, 0) + (0, 6). Finally, by (11-1), we have (a, 0)4- (0,6) = (a,6). Hence, (11-12) is established. Real and Imaginary Parts of Complex Numbers. We call a the real part and & the imaginary part of the complex number a 4- bi. If a = and 6^0, a + bi reduces to bi, which is called a pwre imaginary number. If 6 = 0, the complex number a + bi, or al 4- bi, reduces to the complex number al, which may be identified with the real number a. The complex numbers, then, include both the real numbers and the pure imaginary numbers as special cases. Illustrations of various classes of numbers follow : Some real numbers are —2, 5, and y/S. Some pure imaginary numbers are Si, and — y/5i. Some complex numbers are 2 4- Si, and y/S — i. Note that the numbers —2, 5, \/S, Si, and — y/Ei may be put into the standard form a 4- bi and written, respectively, as the complex numbers -2 4- Oi, 5 4- Oi, \/S 4- Oi, 4- Si, and — \/5i. Since = 04- Oi, which may be written briefly as 0, we shall drop the use of bold-face 0; similarly for bold-face 1. Conjugate Complex Numbers. The conjugate of a complex num- ber a 4- bi is defined as a — bi. Likewise, a 4- bi is the conjugate of a — bi. Some pairs of conjugate complex numbers follow: 2i, —2i; 3 4- 5i, S — 5i; and x 4- 2yi, x — 2yi. A real number is its own conjugate. 192 Complex Numbers Sec. 11-2 Powers of t. It is readily seen that i 3 = i 2 • i = — i, i* = i 2 • i 2 = 1, i& = i* . i = ^ £6 = z *4 . z -2 = _^ an( j so on Therefore, successive posi- tive integral powers of i have only four different values, namely, i, —1, — i, and 1 ; these four values are repeated in regular order. Hence, if n is any positive integer, we have in general i 4n = i 4 •=. 1, i4n+l = i, i4n+2 = i 2 = — i, f4»+3 = i* = — i. These relationships afford a simple method for evaluating powers of i, as shown by the following illustrations : i 7 = i 4+3 — i 3 = — i ; i 38 = i 4 * 9+2 = i 2 = -1 ; and i 103 = i*- 26 *- 1 = t. 11-3. OPERATIONS ON COMPLEX NUMBERS IN STANDARD FORM From the definitions given in Sections 11-1 and 11-2, it follows that a 4- bi and c + di can be added, subtracted, multiplied, and divided as if they were real binomials, except that, where i 1 appears, it is replaced by —1. Algebraic Addition and Subtraction. Addition of two complex numbers is effected by adding their real and imaginary parts sepa- rately ; and subtraction is performed by subtracting their real and imaginary parts separately. Thus, in accordance with (11-1) and (11-2), (11-la) (a + bi) + (c + di) = (a + c) + (b + d) i, and (ll-2a) (a + bi) - (c + di) = (a - c) + (b - d) i. For example, (3 + 20 + (4 - 5i) = (3 + 4) + (2 - 5)i = 7 - 3t, and (3 + 20 - (4 - 50 = (3 - 4) + (2 + 5)i = - 1 + 7i. We note that the sum of conjugate complex numbers is a real num- ber, because (a + bi) + (a — bi) = 2a. Also, the difference of two conjugate complex numbers is a pure imaginary number, because (a + bi) - (a - bi) = 2bi. Algebraic Multiplication. To find the product of two complex numbers, multiply them according to the rules of algebra, and replace i 2 by -1 in the result. Thus (a + 60 (c + di) = ac + adi + bci + bdi 2 . Replacing i 2 by —1, we have, in agreement with (11-3), (ll-3a) (a + bi) (c + di) = (ac - bd) + (ad + bc)i. Sec. 11-3 Complex Numbers 193 In many respects the notation a + bi is more convenient than (a, b) . In particular, the former notation makes it easier to remem- ber how to multiply two complex numbers. For example, (3 + 2i) (4 - 5z) = (12 + 10) + (- 15 + 8)t = 22 - 7t. The student should note that the product of two conjugate complex numbers is a non-negative real number, because (a + bi) (a — bi) = a 2 + b 2 . Algebraic Division. To obtain the quotient of two complex num- bers, multiply the numerator and the denominator by the conjugate of the denominator. Thus, if c + di ¥= 0, a + bi __ a + bi c — di __ ac — adi + bci — bdi 2 c + di ~" c + di c — di "" c 2 — d 2 i 2 __ (ac + fed) + (be — ad)z ~ c 2 + d 2 Therefore ' a + 6i _ ac + 6d , 6c — ad . ^Tdi "^Td 2 + 1?+W %m The right member is, of the form A + Bi, and this equation agrees with (11-10). Example 11-1. Reduce - to the form a -f bi. Solution: The conjugate of i is — i. Then i i ( — t) - i 2 +1 Example 11-2. Find the value of 5 + 13i divided by 3 - 2i. 5 4- 13t Solution: Represent the division as -r — -yr > and multiply the numerator and the denominator by 3 + 2i. We get 5 + 13i 3 +2i (15 - 26) + (39 + 10)t _ 11 49 . 3 - 2i ' 3 + 2i "" 9+4 13 + 13 ^ EXERCISE 11-1 In each of the problems from 1 to 12, express the given quantity in the form a + bi and give its conjugate. In working these problems, note that y/ — a 2 = y/a? y/^~l = | o 1 1. 1. V- 16. 2. V~25. 3. - V^. 4. y/-X*. 7. 3 a/2 + 3 y/~2. 5. - V- 3(K7 2 . 8. 4 - 3 V- 16. 6. V- *V« 9. 1 + 2 V" 17 ^. 10. v/^ 2 - V- * 2 . 11. V15 4- y/- 64a 3 6. 12. 3 + V- 32a 2 6 3 . 1 94 Complex Numbers Sec. 1 1-3 In each of the problems from 13 to 26, compute the value of the given expression. 13. i". 14. i 12 . 15. (- i) 13 . 16. - (- i) 17 . 17. i«. 18. -x 70 19. (-i) 236 . 20. (-i) 602 . 21. i 3 i 29 . 22. t 37 i 183 . 23. i 14 - (- i) 18 . 24. i 10 + i 20 + i 30 . 25. i 25 - i 50 4- i 76 - i 100 . 26. i 10 4- i 100 4- z 1000 4- i 10000 . In each of the problems from 27 to 36, find the values of x and y which satisfy the given equation. 27. (2jc, 3?/) = (3, 1). 28. (2x, Sy) = (8, 9). 29. (3x, 5?/) = (18, - 25). 30. (x + ?/), 3x + 1) = (x, 1). 31. (3* + 2y,x + 5?/) = (2?/ + 3, - 19). 32. z + (y - x)i = 1 + 3i. 33. 3z - 6 - (5 - 2t/)i = 0. 34. 2a; - 4yi = 6 - 2si. 35. 3x - 7 = (4 - Sy)i. 36. 3xt - 2yi + 8x + 5y - 12 = 2x + 2?/i + 5?/ + 6 + (x - 2)i. In each of the problems from 37 to 78, perform the indicated operations and reduce to the form a -f- hi. 37. (2, - 3) + (5, 6). 38. (1, - 1) + (3, 2). 39. (1, V§) + Q > - ^) • 40. (1, - 7) - (7, - 3). 41. (2, 3) • (1, 1). 42. (1, ll 2 . 43. (4, - 3) • (4, 3). 44. (0, - l) 3 . 45. (1, 0) • (0, 1) • (1, 1). «.«>,■>, _ «-{-r y r)'- 48. (- \ , - ^) 3 . 49. V~=^ + V^2 - V9. 50. V" 17 ^ - V 17 ^ - 5t 2 . 51. (4 + 3t) 4- (2 - Si). 52. ( - 2 + Si) - ( - 6 - Si). 53. (8 + W) + (5 + 2*). 54. (3 - 2%) - (3 + 5i). 55. ( - 3 + 2t') - (5 - 2t). 56. (6 + 2i) + (3i + V" 17 !). 57. (2i + 3) + (8 - 5 V 77 !). 58. (6 + 3i) 4- (3 - 5i). 59. (2 4- 5i) (6 - 3i). 60. 2i (5 + 3i). 61. (5 + i) (6 + 2i). 62. (3 - 3i) (4t +2). 63. (1 + i V2) (5 -f 2i). 64. (5i - 8) (2t - 4). 65. (5 - 3t) (2 - 4i). 66. (3 - i) + (2 - 5i). 67. (3i + 4) -*- (1 - i). 68. (2 + i)» -s- (5 - f). 69. (2 - Si) + (5 - 4i). 70. (3 - V3i) + (4 + fit)*. 71. (3 - >/32) 2 + (3 + V3i) 2 - (4 - 3t) (i - 6). 72. (6 - 2i) (1 4- t) (1 - 3t) -s- (4 - fit). 73. 1 -5- (6 - 5i). 74. (6 - fit) 4- 1. 75. i + (4 - 3i). 7fi 3 -4i (4 + i) (i - 5) (3 -I- 50 (8 + 6i) ' D- (2 4- i) (3 - 2t) " - (6 - 5t) (2t - 7) * '° - (4 - 7t) (4 4- 6i) ' 79. Prove that complex numbers satisfy the associative and commutative laws of addition and multiplication and the distributive law. 80. Prove that if (a + bi) (c + di) = 0, then a + hi = or c + di = 0. Sec. 11-4 Complex Numbers 195 11-4. GRAPHICAL REPRESENTATION As we have seen, the complex number a + bi determines a definite point P in the plane whose rectangular coordinates are x = a and y = b. Conversely, to every point P in the plane corresponds a com- plex number a + bi for which the values of a and b are the respec- tive rectangular coordinates of P. See Fig. 11-1. In this sys- tem, the real numbers a + Oi are represented by points on the #-axis, which is called the axis of reals. Pure imaginary numbers 4- bi are represented by points on the 2/-axis, which is called the axis of imaginaries. Fig. 11-1. P 3 (a + c,b + d) *»X Fig. 11-2. **x It is more convenient at times to represent the complex number a + bi by the vector drawn from the origin to the point P. The length of the vector is given by the relationship r = V a " + b 2 , and the direction is given by an angle determined from the equations a = r cos and b - r sin 0. In Fig. 11-2 is indicated the graphical addition of the two com- plex numbers a + bi and c + di, which may be represented either by the points P x and P 2 or by the vectors OP 1 and OP^ With^the completion of the parallelogram 0P x PJP 2 , the sum of OP 1 and OP 2 can be represented by the diagonal OP 3 . Thus, either P 3 (a + c, b + d) or OP 3 represents the sum (a + c) + (6 + d)i. Hence, the vector which represents the sum of two complex numbers is the sum of the vectors representing the given numbers. To subtract c + di from a + bi graphically, we merely add a + bi and — c —di. Example 11-3. Add the complex numbers 2 + 3i and 6 -f 2t graphically. Solution: Let Pi (2, 3) represent the number 2 -f 3i and let P 2 (6, 2) represent the number 6 + 2i. Draw OPi and OP 2 in Fig. 11-3, and complete the parallelo- gram OP1P3P2. Then P3_represcnts the sum 8 + 5i* of the complex numbers 2 + 3t and 6 + 2t", and OP 3 represents the sum of the vectors OPi and OP2. 196 Complex Numbers Sec. 11- P 3 (8,5) Fig. 11-3. The difference of two complex numbers may be obtained in the same manner if we apply the relationship (a + bi) — (c + di) = (a + bi) + (— c — di). Let P and Q represent the numbers a + bi and c + di, respectively, in the complex plane, as shown in Fig. 11-4. Then — c —di is repre- sented by Q', which is the reflection of Q through the origin. Let us recall how the difference of two vectors was explained in Section 6-7 arid was represented graphically in Fig. 6-15. If we let the vectors OP, OQ^ and OQ' represent a + bi, c + di, and -c -di, respectively, then OR is the desired vector and R is the point representing the number (a + bi) — (c + di) . 11-5. TRIGONOMETRIC REPRESENTATION Let the complex number a + bi be represented by the radius vector drawn from the origin to the point P. Then the distance \OP\ = r is called the modulus, or^the absolute value, of the complex number ; and the angle 0, which OP makes with the positive #-axis, is called an argument, amplitude, or angle of the complex number. Sec. 11-5 Complex Numbers 197 />(l/2,V3/2) From Fig. 11-1, it is clear that - = cos and - = sin 0, or a = r cos and b = r sin 0. Hence, a + 6i = r cos + (r sin 0)i = r(cos + i sin 0). This last expression is known as the polar form or the trigono- metric form of the given complex number, as contrasted with the standard or rectangular form a + bi. To reduce a given complex number a + bi to the trigonometric form r(cos +i sin 0), we find r and by means of the relation- ships r = -\/a 2 + b 2 , a = r cos 0, and 6 = r sin 0. We have a + 6i = rf- + - i) = r (cos + i sin 0). Example 11-4. Represent the complex number 1 /3 o + o~ 1# graphically, and change the given notation to the trigonometric form. Solution: The point P whose rectangular coordinates are ( - > -^- J represents the number - -f —^- i. Since a = 1/2 and 6 = a/3/2 are both positive, is a first- quadrant angle. We see from Fig. 11-5 that r = 4/-r+7 = l. The angle 6 is determined from the equations cos = 1/2 and sin = \/3/2. In this case we may let 9 = tt/3 or 60°. Hence, ^ + ^ i = 1 (i + ^ i) = cos 60° + i sin 60°. Example 11-5. Express the complex number 1 - i in the trigonometric form. Solution: Here a = 1 and b = - 1. So r = \/2, and is a fourth-quadrant angle We thus have **X determined from cos = — -p and sin = — V2 * V2 1 - i = V2(-4= - A=<) = v^(cos 315° + t sin 315°). EXERCISE 11-2 In each of problems from 1 to 12, represent the complex number and its con- jugate graphically. 1. 3 + 2i. 2. 8 + 2*. 3. 3 + 4i. 4. 2 - 3t\ 5. 3 - 5i. 6. 1 - 1. 7. i. 8. 1. 9. 10. |(2 + V2i). 11. |(1 - v®). 12. 5 + 12i, 1 98 Complex Numbers Sec. 1 1-5 In each of the problems from 13 to 24, perform the indicated operations graph- ically. Then check the result algebraically. 13. (7 - 3t) + (- 4 + i). 14. (2 + 3t) - (4 - 50. 15. (3 - 60 - (4 + 30. 16. (6 - 20 + (6 + 20. 17. (3 4- 20 - (5 - 0. 18. (3 + 40 - ( - 2 - 40. 19. (5 + + (1 - 50. 20. (3 - 20 - (5 - 40. 21. 3 - (1 - 40 - (2 + 0- 22. (V§ + + (1 + a/30 - 7%. 23. 7 - (4 - 20 - (- 2 + < \/S). 24. (4 + 30 - (2 + 30 - (3 + 20. In each of problems from 25 to 36, change the complex number to the trigono- metric form and represent it graphically. 25. 1 + t. 26. - 5. 27. - 3t\ 28. 3 - 3 y/Zi. 29. hi + VSi). 30. 1(1 - \/30. 31. 5 + 12*. 32. ?• 33.6^4t_l^. 84. (3-4f)». 35.^- 36. %=£• 2 — i 3 +% 6 + 5t 2 -M 11-6. MULTIPLICATION AND DIVISION IN TRIGONOMETRIC FORM Let r x (cos a + i sin a) and r 2 (cos J3 + i sin ft) be any two com- plex numbers in trigonometric form. Then, their product is given by the relationship Ti (cos a + i sin a) • T2 (cos ft + i sin j8) = fir2[(cos ol cos ft — sin a sin ft) + t(sin a cos ft + cos a sin ft)] = r^cos (a+ ft) + isin (a+ ft)]. Thus, we have proved that the absolute value of the product of two complex numbers is the product of their absolute values, and an angle of the product is the sum of their angles. The result found for the product of two complex numbers can be extended to the product of three or more complex numbers. The quotient obtained by dividing the complex number fi (cos a + i sin a) by the complex number r 2 (cos ft + i sin ft) is given by the relationship ri (cos a + i sin a) __ ri(cosa + t'sina) cos /3 — z sin ft r2(cos ft + i sin ft) ~~ r2(cos ft + i sin ft) cos ft — i sin ft = — [cos (a — ft) + i sin (a - ft)]. It follows that the absolute value of the quotient of two complex numbers is the quotient of the absolute values, and an angle of the quotient is the difference of their angles. Sec. 11-7 Complex Numbers 199 Example 11-6. Find the product of 2(cos 30° + i sin 30°) and 3(cos 120° + i sin 120°). Solution: By the rule for products in polar form, we have 2(cos 30° + i sin 30°) • 3(cos 120° + t sin 120°) = 2-3 [cop (30° + 120°) + i sin (30° + 120°)] = 6 [cos 150° + i sin 150°] = 6 ( - ^~ + % ) - 3 ( - y/l + i) • 1 f*\ Example 11-7. Find the quotient when 1 — i is divided by ~ + -^— t. Solution: From Examples 11-4 and 11-5 in Section 11-5, 1 - % = \/2(cos 315° + i sin 315°), and Hence, ^ + ~ i = cos 60° + i sin 60°. 1 - i _ ,- (cos 315° + i sin 315°) 1 , V3 . cos 60° + i sin 60° = \/2 [cos (315° - 60°) + i sin (315° - 60°)] = V2 [cos 255° + i sin 255°] = - y/2 (cos 75° + i sin 75°). From a table of trigonometric functions, we find that the result is - V2 [0.2588 + i (0.9659)]. Or, using exact values of cos 75° and sin 75° previously found, we obtain - V2 [^- (V3 - 1) + i^~ ( V3 + 1) J = - \ [(V3 - I) + % (V3 + 1)]. 11-7. DeMOIVRE'S THEOREM If we extend the law of multiplication of the preceding section to n factors, we have [ri(cos 0i + i sin 0i)] [r 2 (cos 02 + i sin #2)] • • • [r„(cos 6 n + i sin n )] = nr 2 ■ • • r n [cos(0i + d 2 + • • + n ) + i sin (0! + 6 2 + • • • + ft.)]. If how we put ri — r 2 = • • ■ = r n = r and 0i = 2 = • • • = 9 n = 0, it fol- lows that [r(cos + ^ sin 0)] n = r n (cos nd + i sin n0). This result is known as De Moivre's Theorem. Although we have derived De Moivre's Theorem only for integral values of n, it can be shown to hold for all real values of n, if prop- erly interpreted. 200 Complex Numbers Sec. 11-7 Example 11-8. Find the value of (1 - i) 4 by Dc Moivre's Theorem. Solution: Since 1 — i = y/2 (—7=- j-i) > the polar form of 1 — i is \/2 (cos 315° + i sin 315°). Hence, by De Moivre's Theorem, (1 - i) 4 = [y/2 (cos 315° + i sin 315 )] 4 = ( \/2Y [cos (4 • 315°) + i sin (4 • 315°)] = 4(cos 1260° + i sin 1260°) = 4(cos 180° + i sin 180°) = - 4. Example 11-9. Derive formulas for cos 20 and sin 20 by De Moivre's Theorem. Solution: By De Moivre's Theorem for n = 2, we have cos 26 + i sin 26 = (cos +i sin 0) 2 = cos 2 6 + (2 cos sin 0) t - sin 2 0. The two sides are equal only if the corresponding real and imaginary parts are equal. Hence, cos 26 = cos 2 6 - sin 2 6, and sin 20 = 2 sin cos 0. 11-8. ROOTS OF COMPLEX NUMBERS Let p(cos <£ + isin(j)) be an nth root of the complex number r(cos + i sin 0), where 0° ;g ;g 360°. Then [p(cos <t> + i sin <£)] n = r(cos + i sin 0). By De Moivre's theorem, this leads to (11-13) p n (cos n<t> + i sin n<f>) = r(cos + i sin 0). Our problem now is to find all non-negative numbers p and all angles <f> for which (11-13) is satisfied. Separating real and imaginary parts, we have (11-14) p n cos n<t> = r cos 0, p n sin n<£ = r sin 0. Squaring and adding, we obtain p 2n (cos 2 n<£ + sin 2 n<f>) = r 2 (cos 2 + sin 2 0). Therefore, p 2n = r 2 , since cos 2 a + sin 2 a = 1. The absolute value p is then given by the equation (11-15) p = y/r. From (11-14) we then have cos n4> = cos 0, sin n<£ — sin 0. It is clear from these equations that the angles ruf) and can differ only by a multiple of 2tt or 360°. More precisely, (11-16) n<f> = + fc • 360°, or = - + ^^! , where 'A: is any integer. Sec. 11-8 Complex Numbers 201 For fc = 0, 1, 2, ••• , (n-1) in (11-16), we obtain n distinct values of the angle, all of which are non-negative and less than 2rr or 360°. Corresponding to these angles we obtain n distinct roots given by the formula (11-17) S^[ + fc.36O° f . . + fc.36O°- cos 1_ % Sln _ For example, for k — 0, we obtain one nth root, called the prin- cipal root, with absolute value r 1/n and angle - : for k = 1, we have n ft -i— *%f\c\ a second root, with absolute value r 1/n and angle — — ; and so on to k = n — 1. The value k = n would yield the same root as k = 0, 6 + n • 360° n since cos = - + 360°, and 71 e (- + 360°) = cos - and sin in (- + 360°) = sin - • Similarly, k — n + 1 yields the same root as k — 1, and so on. This means that only n distinct roots exist. It is interesting to note that the points which represent the roots are equally spaced on a circle whose radius is ^/r and whose cen- ter is the origin. This fact is illus- trated in the following example and Fig. 11-6. Fig. 11-6. Example 11-10. Find the three cube roots of 8(cos 60° + i sin 60°). Solution: The three cube roots are found by evaluating (11-17). Thus, we have >^8(cos 60° + i sin 60°) = -^(cos 60° + k • 360° + i sin 60° + k • 360° 3 ' 3 / = 2[cos(20° + k • 120°) + i sin (20° + k • 120°)]. As just shown, the substitution of k = 0, 1, 2 yields the three required roots. Hence, for k = 0, we have 2(cos 20° + i sin 20°) ; for k = 1, the root is 2(cos 140° + isin 140°) ; and for k = 2, the root is 2 (cos 260° + i sin 260°). The roots are repre- sented by the equally spaced vectors OP i} OP 2 , and OP 3 , terminating on the circle whose radius is 2 and making angles of 20°, 140°*, and 260°, respectively, with the positive x-axis. 202 Complex Numbers Sec. 11-8 EXERCISE 11-3 In each of the problems from 1 to 18, perform the indicated operations by first expressing the complex number in polar form. Express the answer in rectangular form. 1. (1 + (1 - V3i). 2. (- 1 + t) (V§ + i). 1 -i 3. (-1 + V3i)(\/3 +i). 1 + y/Zi K i - \/3 c 2 + 2i 1 + * V3 + 3t 7 (1 -t)( z l + V&) , 8 (-1 +*)(a/5 V3 + 1 * * + * 11 / I ~ V&V 19 1 +* "' V 2 / 1Z ' (2 + i) (3 + i) i3. „ .\rj. ,. - 14. 1 - v5i (1 + i) (3 + 4i) " ( 2 + 3i) (V3 -f i) (2 - 5 i) 2 (1 + 3i) (3i) 3 1 /0A 70 17. (y/2 - V2t) 10 . 18. 19. [|(1 - V2i)] 100 ' 20. ( - i + i V3t) In each of the problems from 22 to 29, find roots as directed and represent them graphically. 22. Find two distinct square roots of 9 (cos 50° + i sin 50°). 23. Find four distinct fourth roots of 16(cos 36° + i sin 36°). 24. Find three distinct cube roots of 27(cos 165° + % sin 165°). 25. Find five distinct fifth roots of - 32. 26. Find the three cube roots of 1. 27. Find the four fourth roots of 1. 28. Find the two square roots of i. 29. Find the three cube roots of - •= (1 + \/3i). Sec. 11-8 Complex Numbers 203 In each of the problems from 30 to 34, the complex numbers E f I, and Z designate voltage, current, and impedance, respectively, and E = IZ. 30. Compute E when / = 5 + 4i amperes and Z = 30 — Si ohms. 31. Compute J when E = 110 + 30i volts and Z = 20 - 15i ohms. 32. Compute Z when / = 4 + 3t" amperes and ^ = 115 volts. 33. When two impedances Z\ and Z 2 are connected in parallel, the equation -=■=■=-+■=- determines an equivalent impedance Z. Compute Z when Z\ = 5 + 4i ohms and Z* = 8 - 6i ohms. 34. If z and £ are conjugate complex numbers, prove that |*+s| 2 +|z-z| 2 =4|s| 2 . 12 Equations in Quadratic Form 12-1. QUADRATIC EQUATIONS IN ONE UNKNOWN This chapter provides an extension of the work on linear equa- tions to second-degree, or quadratic, equations. Consider a quad- ratic equation in one unknown written in the form (12-1) ax 2 + bx + c = (a ^ 0), where a, b, and c are given real numbers. This equation is called the general quadratic equation in x, and is said to be in standard form. If b ¥" 0, (12-1) is called a complete quadratic equation; if b = 0, it is called a jmre quadratic equation. Thus, 3#~ — a; + 4 — is a complete quadratic equation in which a = 3, 6 = — 1, and c — 4; and # 2 — 2 = is a pure quadratic with a = 1 and c = —2. In Section 12-4 we shall prove that every quadratic equation has two and only two solutions or roots. The roots may be equal or unequal, and they may be real or complex. Their natures depend on the values of a, b, and c. We shall consider the methods in gen- eral use for finding these roots, and we shall then apply them as well to the solution of equations which are not quadratic in x but which can be written as quadratic equations in expressions involv- ing the unknown. 12-2. SOLUTION OF QUADRATIC EQUATIONS BY FACTORING If the left side of a quadratic equation in standard form can be factored, the solution of the equation depends on the following important principle : The product of two or more numbers equals zero if and only if at least one of the factors is equal to zero. That is, A • B = if and only if A = or B = 0. Sec. 12-2 Equations in Quadratic Form 205 In practice, we apply this principle by equating to zero each linear factor of the left side of the given quadratic equation, and solving the resulting linear equations. The following examples will illustrate its application. Example 12-1. Solve 2x 2 - 7x + 6 = by factoring. Solution: To find the values of x which satisfy the equation 2x 2 — 7x + 6 = 0, write the left side in the factored form (x - 2) (2x - 3) = 0. This product equals zero if and only if either x - 2 = or 2x - 3 = 0. Hence, x = 2 or x = 3/2. Moreover, 2 and 3/2 are solutions, because both 2 and 3/2 satisfy 2x 2 - 7x + 6 = 0. Thus, 2(2)2 _ 7(2) +6=8- 14 +6=0, and 2(3/2) 2 - 7(3/2) +6=9/2- 21/2 +6=0. Example 12-2. a) Solve the equation 2 sin 2 x — sin x — 1 = for sin x. b) Find all non-negative angles x less than 360° which satisfy this equation. Solution: a) Factor the given equation to obtain (sin x - 1) (2 sin x + 1) = 0. Since sin x — 1 = or 2 sin x + 1 = 0, it follows that sin x = 1 or sin x = — 1/2. Chech: 2(1)2 - (l) - l = 2 - 1 - 1 = 0, and <-!)'- (-!)->=!+!- — 0. 6) When sin .r = 1, x = 90°; when sin x = - 1/2, s = 210° or 330°. Therefore, j = 90° or 210° or 330°. Check: 2 sin 2 90° - sin 90° - 1 = 2 - 1 - 1 = 0, 2 sin2 210° - sin 210° - 1 = 2( - ^f- ( - ^) - 1 = 0, 2 sin2 330° - sin 330° - 1 = 2^ - i) 2 - ( - ^) - 1 = 0. and Note that this equation is a quadratic in which the unknown is a trigonometric function of the angle x. We thus have only two values of sin x which satisfy the equation. The determination of the angle, however, goes beyond the algebraic solution of the quadratic, and it may happen that there are more than two values of x which satisfy the equation. For this reason, it is recommended that all solutions be checked by substituting in the original equation. 206 Equations in Quadratic Form Sec. 12-2 Example 12-3. Solve the equation 3 sec x = 2 cos x - 1 for all non-negative values of x less than 2ir radians. Solution: Since sec x = > we can write = 2 cos x — 1. We then clear cos x cos x of fractions, transpose, and factor, to obtain 2 cos 2 x — cos x — 3 = (2 cos x — 3) (cos £ + 1 ) =0. Hence, cos x = 3/2 or cos a; = - 1. When cos x = - 1, x = 7r. There is no real number £ for which cos £ = 3/2. 3 sec t = 2 cos tt - 1, or 3(- 1) = 2(- 1) - 1. It should be noted that factoring provides a method of solving any pure quadratic equation ax 2 + c = 0. Thus, the equation is equivalent to which gives x* + £ = L + kT^\ (x - 4/—^) = 0, # + 4/ =0 or # — 4/ = 0, " a V a or # = ± 4/ • This result agrees with that given by writing x 2 = f (X tZ and then simply extracting square roots of both sides to obtain / — c c x = zb 4/ • Note that the roots are real when -^0 and pure imaginary V a a when - > 0. a EXERCISE 12-1 Solve each of the following equations for x or 0. In each of the problems from 9 to 20, find all non-negative angles less than 360° which satisfy the given equation. Check all solutions. 1. x 2 + 7x = 0. 2. x 2 - lOx +21 =0. 3. 2x(Ax + 5) = - 3. 4. x 2 + 6.r - 27 = 0. 5. x 2 - Z = 6. 6. 6x 2 - 4x - 192 = 0. 7. x 3 + 27 - 3x(x +3) =0. 8. 16z 2 - a 2 + 2a6 - b 2 = 0. 9. sin 2 - sin = 0. 10. sin = esc 0. 11. sin + 1 = 2 esc 0. 12. 2 tan 2 + 3 tan - 2 = 0. 13. sec (sec + 6) = 16. 14. 3 cos 2 - 8 sin = 0. 15. (cot - l).(cot + 2) = 4. 16. ?^i * = J2 _ _4 # v 3 6 esc 2 esc 17 cot + 4 143 1ft 8 2+3 cos _ cot + 6 ~ (cot + 6) 2 3 cos + 3 ^ 3 cos + 4 3 7 _ 5 4 cos 2 8 cos 2 19. 3 sin 2 0=4 sin 3 0-10 sin 0. 20. , 3 - „ - 77-^ - | = 0. 12-3. COMPLETING THE SQUARE The method developed here is based on the fact that we can make any binomial of the form x 2 + kx into a perfect square if we add to Sec. 1 2-3 Equations in Quadratic Form 207 it the square of one-half the coefficient of x. To make this clear, let us recall from Section 1-18 the formula for a perfect-square trinomial. The formula is (x + a) 2 = x 2 + 2ax + a 2 . Since the coefficient of x in x 2 4- kx is k, the square of one-half of /k\ 2 k 2 the coefficient is I = ) or -j • Adding this to # 2 + &#, we have x 2 + kx + -^ = (a; + ~j • Thus, the left member is a perfect square, namely, the square of x + r Applicability of the procedure to a variety of processes, including solution of quadratic equations, is illustrated in the following examples. Example 12-4. Solve x 2 — 2x — 4 = by completing the square. Solution: We first transpose the constant term, so that the left side will be of the form x 2 + kx. Hence, the equation becomes x 2 - 2x = 4. Now the quantity ( — l) 2 = 1 is added to the left side to make it a perfect square. To obtain an equivalent equation, the same quantity is added to the right side also. The result is x 2 - 2x + 1 = 5, or (x - l) 2 = 5. Taking square roots of both sides, we have x - 1 = ± y/b. So the desired solutions are x = 1 + \/5 and x = 1 — \/h. Check: j (1 + y/l) 2 - 2(1 4- a/5) -4 = 1-1-2 V5 +5-2-2 >/5 -4=0, and _ _ _ (1 - V5) 2 - 2(1 - V5) - 4 = 1 - 2 V5 + 5 - 2 + 2 y/b - 4 = 0. Example 12-5. Solve 2x 2 - 5x + 3 = by completing the square. Solution: Transpose the constant term to obtain 2x 2 - 5x = - 3. Since the coefficient of x 2 is not 1, we make it 1 by dividing both sides by 2. Then we have 2 5 3 * 2* = ~2* 208 Equations in Quadratic Form Sec. 12-3 The square of half the coefficient of x is (^ ( ~~ o) ) = Tp ' ^d tm s numDer to both sides, thus making the left side a perfect square. The result is 2 5 , 25 _ 3 25_J_ * ~2* + 16~ ~2 + 16~10* 5 1 When we take square roots of both sides, we have x — j = = t • Solving for z, we obtain That is, Check: and 5 , 1 # = « or x = 1. 2(1) 2 -5(1) +3=2-5+3=0. Example 12-6. Reduce .r 2 + 2/ 2 - 4x + 62/ + 4 = to the form (a: - /i) 2 + (1/ - A;) 2 = r 2 . Solution: The solution of this problem requires that we complete the square of the terms containing y as well as the square of those containing x. Hence, for convenience, we write the equation in the form Or 2 - 4x ) + (y 2 + %y ) = - 4. When we complete the squares in the parentheses, the equation becomes (x 2 - Ax + 4) + (y 2 + 6// + 9) = - 4 + 4 + 9. Thus, the solution is (a? - 2) 2 + (y + 3) 2 = 9. Example 12-7. Reduce 9x 2 - Ay 2 - I8x - 16?/ - 43 = to the form A(x -h) 2 - B(y -k) 2 = C. Solution: Write the equation in the form 9(x 2 - 2x ) - A(y 2 +4?/ ) = 43. Complete the squares in the parentheses to obtain 9(x 2 - 2x + 1) - A(y 2 + 4z/ + 4) = 43 + 9 - 16. Note that the numbers 1 and 4, which are added within the parentheses to complete the squares, must be multiplied by the coefficients 9 and - 4, respectively, to determine the numbers that are added to the right side. The reduced form is, therefore, 9(3 - 1)2 _ 4 (j, + 2)2 = 36. Example 12-8. Reduce y/Zx 2 + 4a: - 4 to the form y/a{(x - h) 2 - k 2 ). Solution: For convenience, work with the quantity 3x 2 + Ax — 4 without the radical sign until the final result is obtained. Hence, write 3z 2 + Ax - 4 = %(x 2 + |r - |) • Sec. 12-4 Equations in Quadratic Form 209 Complete the square of the terms in x and simplify to obtain »(-+J-J)-»((-^+S)-S-!)=»(('+i) , -¥)- Now, write this result under the radical sign to obtain 4/3M x +^J — 7j") * Comparing this with the required form <\/a((x — h) 2 — k 2 ), we see that we may choose a = 3, h = - 2/3, and k = d= 4/3. EXERCISE 12-2 In each of problems from 1 to 15, solve the given equation by completing the square. In each of the problems from 9 to 15, find all non-negative angles less than 300° which satisfy the given equation. Check all solutions. 1. x 2 - Sx = 20. 2. x 2 + lOx = 40. 3. x 2 - 7x = 30. 4. z 2 + £ + 1 = 0. 5. x 2 + x + 2 = 0. 6. 6x 2 - 5x - 1 = 0. 7. x a + s - 5 = 0. 8. 2z 2 = 3x + 9. 9. tan 2 = 2 tan + 1. 10. cos 2 = -^+3. 11. ^o=^- sec; sec + 2 3 cop 4-9 12. 1 + tan-' = sec 0+3. 13. —^n~ = 2. esc 2 0—1 14. 3 cot 2 + cot = 3 - 4 cot 0. 15. sec - cos = 2. In each of the problems from 1(5 to 25, reduce the equation to the form A(x - h) 2 + B(y - k) 2 = C. 16. x 2 - Ay 2 - 2x + 1 = 0. 17. x 2 + 4i/ 2 - 6z + 16?/ + 21 = 0. 18. 4x 2 + 9?/ 2 + 32.Z - lSy + 37 = 0. 19. x 2 + 4// 2 - 10-r - 40?/ + 109 = 0. 20. 9x 2 + 4?y 2 - 8y - 32 = 0. 21. 4* a + 9</ 2 - lfxc - 18/y -11=0. 22. x 2 - 9?/ 2 - 4x 4- 36;// - 41 = 0. 23. 4x 2 - 9y 2 + 32x -f 36*/ + 64 = 0. 24. by 2 - 4r 2 + 50*/ -f- 32x + 41 = 0. 25. 3x* - ?/ 2 + 20x - 2y + 11 = 0. In each of the following problems, express the quantity inside the radical or parentheses in the form a((x — h) 2 ± A; 2 ). 26. V* 2 - 6x - 40. 27. v/2x» - 16 j + 41. 28. (4// 2 - 20y - 76) 3 ' 2 . 29. — 1 = • 30. Or 2 - to + 34) 2 ' 3 . 31. (2x 2 + 28.c + 34)" 1 ' 2 . V# 2 — 6x — 7 32. V(9x 2 + 48.x + 23) 3 . 33. (9x 2 + 24x + 25)" 1 ' 3 . 34. (7.r 2 - 14z + 11)~ 3 . 12-4. SOLUTION OF QUADRATIC EQUATIONS BY THE QUADRATIC FORMULA By applying the method of completing the square to the general quadratic equation (12-1), we can obtain a formula for the roots,, either real or complex, of any quadratic equation whatever. The general equation is (12-1) ax 2 + bx + c = 0. Transpose the constant term c, and obtain ax 2 + bx = — c. 2 1 Equations in Quadratic Form Sec. 1 2-4 Dividing by a, we have 2 f o c X * + —x = • a a /l 6\ 2 ft 2 Add (?:•-) = t~9 to both sides to obtain \2 a/ 4a 2 o . ft , b 2 b 2 c a 4a 2 4a 2 a which becomes , , % «, , 2 ,, Extracting square roots of both sides gives ■ ft _ =1= \/b 2 - 4ac * + 2a ~ 2a Solving for x, we have — ft ± s/b 2 — 4ac * = £ • Hence, to solve a quadratic equation, put it into the standard form ax 2 + bx + c = 0, and substitute the coefficients a, b, and c in the formula just derived to obtain the roots (12-2)si= ^ and z 2 = ^ That these numbers Xj. and #2 are solutions of the given quad- ratic equation is shown by substituting each of them in (12-1). The details of this substitution for Xi follow : aXl 2 + bxi + c = ^ _ J + ^ Ya j + c - / ft 2 ~ 2b Vb 2 - 4oc + (*> 2 - 4ac) \ 4a 2 / = «(- b 2 + b y/ tf—lac + _ +c _ b 2 - 2ac - b Vb 2 - 4ac , - 6 2 + & V& 2 - 4ac , _ 2a" + 2a + C Hence, the number # x satisfies the equation a# 2 + &# + c = 0. A similar computation shows that x 2 is also a solution of a# 2 + bx + c = 0. Consequently, the two expressions given for # in (12-2) are actually roots of (12-1). In Section 12-7 we shall study the expres- sions in (12-2) further and shall determine when the roots are dis- tinct and when they are real. Sec. 12-4 Equations in Quadratic Form 21 1 Example 12-9. Solve 5x 2 - 6# - 8 = by the quadratic formula. Solution: Here a = 5, fr = — 6, c = - 8. Substituting these values in the formula, we obtain ~(-6)±V(-6)2-(4)(5)(-8) x _ _ _ 6 ± VS6 ± 160 6 ± \Zl96 6 db 14 10 10 10 tu f 8 + 14 , 6-14 4 Therefore, Zi = — r^r — = 2 and x 2 = — jg — = - r • These values are seen to satisfy the original equation when substituted for x. Example 12-10. Solve x 2 - x + 2 = by the formula. Solution: Since a = 1, b =■ — 1, c = 2, we have _ 1 d= yT^"8 _ 1 db \/^T _ 1 db \/7i m X "" 9 "" 9 "" 9 Hence, JCl = -Z an( J ^ = « Check: ( I ±V7i \* 1 +VTi , _ -6+2V?i 1 + \/7j , V 2 / 2 + 2 ~ 4 2 + J = -6+2\/7i-2-2V7i+8 :=0 4 Similarly, the second root may be checked. Example 12-11. Solve 2 sin 2 x + 3 cos x - 3 = by the formula, determining all non-negative angles x less than 360°. Solution: We make use of the identity sin 2 x + cos 2 x = 1 to transform the given equation into one involving a single trigonometric function of x> Replacing sin 2 x by 1 — cos 2 x, we have 2(1 - cos 2 x) + 3 cos x - 3 = 0. Simplifying, we obtain 2 cos 2 x — 3 cos x -f 1 = 0. Solving for cos x by the formula, we find 3 ± V9 -8 3±1 cos a; = ^ = — z — • 4 4 Hence, cos x = 1 or 1/2. Therefore, x = 0° or 60° or 300°. The solutions may be checked by substitution in the original equation. Example 12-12. Solve cos x tan x + sin 2 x = 1 - sin x by the formula, deter- mining all non-negative values of x less than 360°. sin x Solution: Making use of the identitv tan x = > we have cos x sin x . . _ i cos £ . ^ sm 2 x = i — sin 3. cos 3 2 1 2 Equations in Quadratic Form Sec. 1 2-4 Transposing and simplifying, we get sin 2 x + 2 sin x — 1 = 0. Solving for sin x by the quadratic formula, we obtain - 2 ± 2 V2 - /o sin z = ^ — = - 1 ± v 2. The value sin x = — 1 + V2 is one solution. However, sin x = — 1 — \/2must be excluded, since the sine of an angle cannot be numerically greater than 1. Check: For sin x = \/2 — 1, we find that V2 - 1 cos x = \/2V2 — 2 and tan # = - V2V2 - 2 Substituting these values in the original equation, we have V2V2-2- v/2 "~ 1 + (V2 - 1)» = 1 - (V2 - 1). V2\/2 - 2 This reduces to V2 - 1 + 3 - 2\/2 = 1 - V2 + 1 or 2 - V2 = 2 - V2. From the table of trigonometric functions, we have x = 24°28' or 1 55°32'. EXERCISE 12-3 Solve each of the following equations for x or by the quadratic formula. In each of the problems from 14 to 24, find all non-negative angles less than 360° which satisfy the equation. Check all solutions. 1. s a + 6x - 7 = 0. 2. x 2 - x - 20 = 0. 3. x 2 + 2z - 5 = 0. 4. .t 2 + 3 + 1 = 0. 5. x 2 + 2x + 1 = 0. 6. 7.r 2 - Sx - 9 = 0. 7. 2x 2 + 3.z + 2 = 0. 8. <Xr 2 - 7x - 5 = 0. 9. (2x - l) 2 - 2(2.c - 1) - 8 = 0. 10. (a 2 - b 2 )x 2 - iabx - (a 2 - b 2 ) = 0. 1L J5 _J_ =4 . 12. -A- + 8 - 3 x+2 2x+3 ' &-3 1-3B+1 15. --^ ~ = 3. 16. esc - 3 = sin 0. tan cot 17. sin 2 - 2 sin = sin + 3. 18. 16 sec 2 - 8 sec + 1 = 0. 19 5 , _J_ . 3 - 20 cot * + 2 - 2 cot - 1 _ 2 "^ cos "*" 2 cos 2 2 cot - 3 cot ~ 21. (sec + 1) (sec + 2) = sec + 3. 22. (sin + 3) (3 - sin 0) = 3(sin + 3). 23. (esc + 2) 2 + esc = 1. 24. — l^-r + J = 0. cos — 1 Z 12-5. EQUATIONS INVOLVING RADICALS Sometimes an equation in which the unknown appears under a radical sign can be reduced to a quadratic by raising both sides to Sec. 12-5 Equations in Quadratic Form 213 a power sufficient to remove the radical. The process must be repeated until the unknown no longer occurs under a radical. The operation of raising both sides of an equation to a power may lead to an equation redundant with respect to the original ; that is, the final equation may possess roots that are not roots of the orig- inal equation. Such roots are called extraneous roots. For this reason, every root obtained must be checked by substitution. Example 12-13. Solve the equation y/x 2 - 3a; + 4 = 2. Solution: Cube both sides to obtain x 2 - Sx + 4 = 8. Transpose, and get x 2 - Sx - 4 = 0. Factoring and solving for x, we find that x = — 1 or x = 4. Check: and ^ ( ~ 1)2 " 3( ~ 1} +± = < /i + 3 + 4= V8=2, ^(4)2 Z 3(4) + 4 = j/s = 2. Hence, both — 1 and 4 are roots. Example 12-14. Solve the equation y/2x - 1 — V# +3 = 1. Solution: Transpose one radical to obtain y/2x - 1 = 1 + y/x +3. When both sides are squared, the result is 2x - 1 = 1 + 2aA +3 + z + 3. Combining like terms, we obtain a: - 5 = 2aA + 3. Now we square both sides to get x 2 - 10* + 25 = 4(a -I- 3). Transposing and combining gives :r 2 - 14s + 13 = 0. By factoring and solving, we find that x = 1 or x = 13. Check: and V2(l) -1- VI +3 = 1-2*1, V2(13) - 1 - V13 +3 =5-4 = 1. Hence, 13 is a root, but 1 is not. EXERCISE 12-4 Solve each of the following equations. In each case check for extraneous roots 1. y/x ~ 2 = 4. 2. V3z +4 = 2. 3. Vz + 5 = 1. 4. V3z -1=7. 5. y/x 2 - 16 = 2-r 1 '*. 6. V* 2 - 2 = V2.r + 0. 214 Equations in Quadratic Form Sec. 12—5 7. x - 3 - y/xT^l = 0. 8. x - 5s 1 ' 2 +6=0. 9. V2x + 3 = 4 - 3x. 10. \/z - 1 + \/x - 3 = 2. 11. \/5 - x + \/4x™T~5 = 5. 12. V2z + <\/2x - 4 = 2. 12-6. EQUATIONS IN QUADRATIC FORM Frequently, an equation which is not quadratic in the given unknown may be considered as a quadratic in some expression involving the unknown. Thus, ar 4 — 3ar 2 + 2 = and 2(x 2 — 2x) 2 — (x 2 — 2x) — 6 = may be treated as quadratic equations in the expressions ar 2 and (x 2 — 2x), respectively. This type of situation was met earlier in Examples 12-11 and 12-12. The following examples will further illustrate methods used in solving equations in quadratic form. Example 12-15. Solve the equation x~* - 3x~ 2 +2=0. Solution: Let x~ 2 = y, so that the given equation becomes y 2 - Sy + 2 = 0. Factor, to obtain (y - l) (y - 2) = 0. Therefore, y = l or y = 2, or ar» =1 or x~ 2 = 2. Hence, the solutions are x = =fc 1 and a; = ± — 7= • V2 Example 12-16. Solve (x 2 - 2) 2 - 7(.r 2 - 2) + 10 = 0. Solution: Let x 2 — 2 = 2/, so that we get ?/ 2 - ly + 10 = 0. Factor, to obtain (V - 2) (y - 5) = 0. Hence, 2/ = 2 or y = 5, or x 2 - 2 = 2 or .x 2 - 2 = 5. Then, x 2 = 4 or x 2 = 7, and the roots are & = ± 2 and x = ± y/l. Sec. 12-7 Equations in Quadratic Form 215 Example 12-17. Solve x 2 + x - 2 y/x 2 + x + 3 = 0. Solution: Let V^ 2 + x + 3 = ?/. Then we can write (z 2 + z + 3) - 2 -\A 2 + z+3-3=0, or y 2 - 2?/ - 3 = 0. Therefore, y = - 1 or 3, and y/x 2 + z+3=-lor y/x 2 + x + 3 = 3. By definition of the radical, y/a is a non-nogative number. Hence, although y/x 2 +z+3=— lis consistent with the original equation, there are no values of x which satisfy this equation. Consider, then, y/x 2 + x + 3 = 3. This leads to x 2 + x + 3 - 9, or x 2 + z - 6 = 0. Hence, x = 2 or a = — 3. Substitution shows that each of these values of x satisfies the original equation. EXERCISE 12-5 Solve each of the following equations. Check all solutions. 1. x* + x 2 - 12 = 0. 2. 4x~ 4 - llx- 2 -3=0. 3. {x 2 + 2) 2 + Six 2 + 2) - 4 = 0. 4. x* - 6x 2 + 8 = 0. 5. z 4 - 13x 2 + 36 = 0. 6. x 4 - 1 = 0. 7. (3x - 4) 2 + 6(3* - 4) + 13 = 0. 8. (x 2 + 3z) 2 - 14(x 2 + 3x) + 45 = 0. 9. (x + -Y+ 2(x + -) - 48 = 0. 10. (x 2 - x) 2 - 20(x 2 - x) + 36 = 0. 12-7. THE DISCRIMINANT It will be recalled that the two roots of the general quadratic equation ax- + bx 4- c = are no on -b + y/b 2 -An~c -b- y /b 2 - lac (12-2) «!= ^ and * 2 = 2a The expression 6 2 — 4ac, which appears under the radical sign, is called the discriminant of the quadratic polynomial ax 2 + bx + c, or the discriminant of equation (12-1). In what follows we shall assume that a, b, and c are real, and we shall make use of the discriminant to determine the character of the roots without actually solving the equation. By inspection of the solutions x x and x 2 , we reach the following conclusions : 1. If b 2 — 4ac = 0, each of the two roots x x and x 2 is equal to — — , and both roots are thus real. 2a 216 Equations in Quadratic Form Sec, 12-7 2. If 6 2 — 4ac is positive, then y/b 2 — Aac is real, both roots are real, and they are distinct. 3. If b 2 — 4ac is negative, then \/b 2 — 4ac is imaginary, and the roots are distinct complex numbers of the form a + fii and a — f3i. These results may be summarized as follows : Value of Discriminant Character of Roots >0 = <0 Real and unequal Real and equal Unequal conjugate complex numbers Furthermore, if a, b, and c are rational, and b 2 — 4ac is a perfect rational square, then the roots are rational; otherwise, they are irrational. The following examples will illustrate how to determine the character of the roots. Example 12-18. Determine the character of the roots of 2x 2 + 7x - 15 = 0. Solution: Here a = 2, b = 7, c = — 15. Hence, 62 - 4ac = 49 + 120 = 169 = (13) 2 . The discriminant is positive, and so the roots are real and unequal. Since the discriminant is a perfect square, the roots are also rational. Example 12-19. Determine all values of k for which the roots of the equation kx 2 - 2kx +4=0 are equal. Solution: The discriminant must equal zero for the equation to have equal roots. Hence, b 2 - 4ac = U 2 - 16& = 4k{k - 4) = 0, and so k - or k = 4. When k = 0, the equation is not quadratic. Therefore, the roots are equal only when k = 4. EXERCISE 12-6 Determine the character of the roots of each of the following equations by means of the discriminant. 1. x 2 + 3x + 4 = 0. 2. x 2 + &c - 9 = 0. 3. x 2 + 16x - 6 = 0. 4. 6z 2 - 7x + 3 = 0. 5. z 2 + lOz + 2 = 0. 6. x 2 + 3x + 2 = 0. 7. 5z 2 + 7x + 2 = 0. 8. 4x 2 - I2x + 9 = 0. 9. x 2 - 2x + 1 = 0. 10. 4x 2 - Sx + 2 = 0. 11. 2x 2 - re + 3 = 0. 12. 5x 2 + & + 5 = 0. 13. 4x 2 - I2x - 9 = 0. 14. x 2 4 4x - 18 = 0. 15. x 2 - 4x + 7 = 0. Sec. 12-8 Equations in Quadratic Form 217 12-8. SUM AND PRODUCT OF THE ROOTS Adding the two roots of the general quadratic equation ax 2 + bx + c = 0, we obtain, by using (12-2), » - ft + v^ 2 E 4ac -l ~ b z v 62 E 4ac -_?&-_ ^ ^ + * 2 - 2a + 2a ~~ 2a ~ a ' Also, multiplying these roots, we have - b + y/b 2 - Aac - b - \/& 2 - 4ac XlX2 = 2a 2a — & 2 — (b 2 — 4ac) __ 4ac _ c "" 4a 2 ~~ 4a 2 ~~ a Hence, we have, for the sum and product of the roots, (12-3) xi + x 2 = ~ - > and (12-4) £i£2 = - • These formulas are used in various ways, for example, in check- ing roots of a quadratic equation, and in forming an equation if its roots are known. To find the factored form of the quadratic polynomial ax 2 + bx 4- c, let us make use of the sum and product formulas just found. Solving #i + x 2 = for b, and solving X\X 2 = - for c, we have a a b = — a(x x + x< 2 ) and c = a(x^x 2 ). Hence, by substitution, we have ax 2 + bx + c = ax 2 — a(x\ + #2)x + a(zi£2) = a(x 2 — (#i + xi)x + (#1X2)) = a(x — x{) (x — #2). Therefore, if x A and :r 2 are the roots of the quadratic equation ax 2 + bx + c = 0, its factored form can be written a(x — x{) (x — X2) = 0. Since a^O, we may write the equation equivalently as (x — #i) (x — #2) = 0. This form or its expansion, X 2 — (Xi + X2)X + X1X2 = 0, may be used in writing an equation whose roots are known. Example 12-21 indicates the procedure. Example 12-20. Without solving the equation, find the sum and the product of the roots of the equation Sx 2 — 5.c +2=0. Solution: In this case, a = 3, b = - 5, and c = 2. Then the sum of the roots is = - , and the product of the roots is - =77' a 3 a 3 218 Equations in Quadratic Form Sec. 12-8 Example 12-21. Write a quadratic equation in the form (12-1), given that the roots are (1 ± y/3 i). Solution: Since x x = 1 + V3 i and x 2 = 1 — s/S i, we have and Xl + X ' 2 = (1 + ^ ^ + (1 " ^ ^ = 2j Zi.r 2 = (1 + V3 i) (1 - \/3 i) = 1 - 3 i 2 = 4. Therefore a suitable equation is x 2 - 2 c + 4 = 0. Alternate Solution: Writing the desired equation in factored form, we have (a: - (1 + V3 i)) (a; - (1 - V3 i)) = 0. Simplifying, we obtain (a; - 1 - V3 (a - 1 + V3 i) = 0, ((* - 1) - V§ ((* - 1) + V3 f) = 0. (a; - l) 2 + 3 = 0, and, finally, x 2 - 2x + 4 = 0. or Hence, EXERCISE 12-7 In each of the problems from 1 to 9, find the sum and the product of the roots of the given equation. 1. x 2 + 2x - 1 = 0. 2. 3.r 2 - x + 2 = 0. 3. x 2 + 2 = 0. x% \ x + | = 0. 6. 5* 2 4. 6a: 2 - 2.c + 3 = 0. 2 4 ' 5 7. 5.r 2 - 6a: - 1 = 0. 8. 5x 2 + 6* + 1 = 0. 9. lOOx 2 Form an equation with each of the following pairs of roots. 11. 0, + 2. 15. 2 ± tV5. 12. 16. - 1, 1. (1 ± iV3). 19. 0, V3 - V5. 20. a ± W. 6a; + 1 = 0. - 40x + 17 = 0. 13. 3, 6. 17. ±i. 21. V3 =fc V5 i. :*i*o) '(-6/2o,o) **X 12-9. GRAPHS OF QUADRATIC FUNCTIONS To graph any quadratic function of the form ax' 2 + bx + c, we set y = a# 2 4- 6a; + c and construct a table of values of y corresponding to assigned values of x. The graph is of the type shown in Fig. 12-1 and is called a parabola. As found by the quadratic formula, the two solutions of the general quad- ratic equation (12-1) are given by (12-2) and Fig. 12-1. xi = X 2 = -b + Vb 2 - - 4ac 2a - b - Vb 2 ■ - Aac 2a Sec. 12-9 Equations in Quadratic Form 219 Since (12-1) states that y = in the equation y = ax 2 + bx + c, the solutions Xi and x 2 are ^-intercepts of the curve. In Fig. 12-1, let A and C be the ^-intercept points, and let B be the mid-point of AC. We note that OB = OA + AB, Xi — X 2 Xi + £ 2 OB = x 2 + - T - = —2— = - ^ • Therefore, B is the point f — jr- > J Now consider the equation y = fc, which represents a straight line parallel to the x-axis. This line may or may not intersect the parabola, depending on the value of k. Solving the equations y = k and y = ax- + bx -f c simultaneously, by elimination of y we obtain ax- + bx + c — fc = 0. The roots of this resulting equation are (12-5) xi- -A + and 2a * 2 ~ 2a 2a v^ - 4a(f - -k) 2a V& 2 ■ - Aa(c - -k) If the value of k is such that y — k intersects the curve in two distinct points, the discriminant in (12-5) is greater than zero, and the roots will be real and distinct. The point on the line y — k with abscissa x = — ^- is then equidistant from the points of inter- section, whose abscissas are x 1 and x 2 . If, on the other hand, the value of k is such that y — k does not intersect the curve, the discriminant in (12-5) is less than zero, and we have a pair of conjugate complex roots, In this case, — — is the real part of these roots. The points with abscissa x = — — and arbitrary ordinates lie on a vertical line, called the axis of symmetry, or simply the axis of the curve; the curve is said to be symmetric with respect to the axis. The point of intersection of the axis and the parabola is called the vertex of the parabola. If the coefficient a of the second-degree term of y = ax 2 + bx + c is positive, the vertex is the loivest point, and the curve is said to be concave upward. If a is negative, the 220 Equations in Quadratic Form Sec. 12-9 vertex is the highest point, and the curve is said to be concave dotvmvard. To find the coordinates of the vertex, we solve simultaneously the equation x = — — of the axis and the equation y Ad ax 2 4- bx + c of the parabola. The coordinates of the vertex are thus found to be (12-6) x = — 2a and 0-^2 2a + C ~ 4a The vertex may be characterized in another way. Let us demand that a horizontal line y = k intersect the parabola in two coincident points. That is, let us insist that the two roots x x and x 2 in (12-5) coincide. Then the discriminant in (12-5) is equal to 0, and x x = x 2 = — o- • The value of y corresponding to this value of x is Ztd, the ordinate of the vertex, as found in (12-6) . We say that the line b 2 — 4ac y = - 4a is tangent to the parabola at the vertex. Example 12-22. Graph x 2 - Gx + 4. Solution: Let i/ = x 2 — 6x + 4, assign values to #, and compute the corres- ponding values of ?/, as in the accompanying table. The graph is shown in Fig. 12-2. The coordinates of the vertex are b 6 n b 2 x= - — = - = 3 and y = c - ~ = 4 2a 2 4a 9 = - 5. C— 1.11) +»x X V - 1 11 4 1 - 1 2 -4 3 - 5 4 -4 5 - 1 6 4 7 11 (2,-4) Hence, the axis is the line x = 3. Since a is positive, the vertex (3, — 5) is the lowest point on the curve, and the curve is concave upward. Sec. 12-10 Equations in Quadratic Form 221 Example 12-23. Graph y = x 2 - 6a; + 9 and y = x 2 - Qx + 14 relative to the same coordinate system as was used for the graph of y = x 2 — 6# -f 4. Solution: Tables similar to that in Example 12-22 but applying to the first two curves are constructed. The three curves are shown in Fig. 12-3. X V = .t 2 - fxc + 9 - 1 16 9 1 4 2 1 3 4 1 5 4 6 9 7 16 X £' = .r 2 - 6x + 14 - 1 21 14 1 9 2 6 3 5 4 6 5 9 6 14 7 21 >x Fig. 12-3. Curve (A) crosses the x-axis at two points, corresponding to the roots 3 =fc \/5 of .r 2 — Gx -\- 4 = 0. Curve (B) is tangent to the .r-axis at (3, 0), because both roots of x 2 — 6.c + 9 = are equal to 3. Curve (0) does not intersect the a>axis, because 1 x 2 — (ir + 14 — has imaginary roots. The reader should relate the discriminants of the quadratics to a study of these graphs. 12-10. QUADRATIC EQUATIONS IN TWO UNKNOWNS The general equation of the second degree in x and y is (12-7) ax 2 + bxy + cy 2 + dx + ey + / = 0, where a, b, c, d, e, and / are given real numbers. An equation of this form, in which at least one of the coefficients a, b, and c is different from zero, is called a quadratic equation in x and y. By a solution of such an equation, we mean a pair of real or complex numbers which, when substituted for x and y in (12-7), will reduce the left side of the equation to zero. Usually there are infinitely many pairs of numbers which satisfy the equation. 222 Equations in Quadratic Form Sec. 12-10 If c ¥" 0, (12-7) may be solved for y in terms of x by means of the quadratic formula. Corresponding to each real value assigned to x, we then obtain, in general, two values of y. We then have pairs of numbers (x, y) which, if real, may be plotted in a rec- tangular-coordinate system. It is shown in analytic geometry that the graph so obtained will be one of a class of curves called conic sections, inasmuch as they may be obtained as curves of intersec- tion of a plane and a right circular cone. At this time we shall confine ourselves to merely listing the curves which comprise this class and indicating briefly the form of the quadratic that corre- sponds to each of the graphs. A more adequate discussion of this subject is given in analytic geometry. However, typical examples of these curves are given here. 1. Parabola. When A^O, the equations y = Ax 1 + Bx + C and x = Ay 2 4- By + C represent parabolas with vertical and horizontal axes of symmetry, respectively. 2. Circle. When C is positive, the equation x 2 + y 2 — C represents a circle whose center is at the origin and whose radius is \A?- 3a. Ellipse. When the constants are positive, the equation Ax 2 + By 2 = C represents a curve called an ellipse. If A = B, the ellipse is a circle. 3b. Point Ellipse. If A and B are positive and C — 0, the equa- tion Ax 2 + By 2 = C is satisfied by only one point, namely, the origin. The graph is then said to be a point ellipse. 3c. Imaginary Ellipse. If A and B are positive and C < 0, there are no (real) points on the graph, and we say that the equation Ax 2 + By 2 = C represents an imaginary ellipse. 4a. Hyperbola. When A, B, and C are positive, the equations Ax 2 — By 2 = C and Ay 2 — Bx 2 — C represent hyperbolas. 4b. Hyperbola. When C ¥= 0, the equation xy — C represents a curve called an equilateral hyperbola. 5. Pair of Straight Lines. The equation Ax 2 + Bxy + Cy 2 + Dx -f Ey + F = represents two straight lines, which may be either distinct or coincident, if the left side can be expressed as the prod- uct of two real linear factors. We use the quantity b 2 — 4ac, which is usually called the charac- teristic of the equation ax 2 + bxy + cy 2 + dx + ey + / = 0, to deter- mine the nature of the conic corresponding to a particular form of the general quadratic equation. In analytic geometry the following statements are shown to be true : 1. If b 2 — 4ac = 0, the conic is a parabola or two real or imagi- nary parallel lines. 2. If b 2 - 4ac < 0, the conic is an ellipse or a point ellipse or an imaginary ellipse. Sec. 12-10 Equations in Quadratic Form 223 3. If b 2 — 4ac > 0, the conic is a hyperbola or two intersecting lines. In Section 12-9 the graph of the parabola y = ax 2 + bx + c was discussed. In the following illustrative examples, the procedures for graphs of other quadratic equations are considered. Example 12-24. Graph x 2 + y 2 = 9. Solution: Set y = to obtain the ^-intercepts, which are ± 3 ; and set x = to obtain the ^-intercepts, which are ± 3. Solve the equation for y, obtaining y = ± V9 - x 2 . Then construct a table of other corresponding values pf x and y. To yield real values of ?/, the numerical value of x cannot exceed 3. As shown in Fig. 12-4, the resulting graph is a circle with center at the origin and radius 3. x -4 -3 -2 - 1 1 2 3 4 y imaginary ±2.24 ±2.83 ±3 ±2.83 ±2.24 imaginary Fig. 12-4. Example 12-25. Graph 4z 2 + 92/ 2 = 36. Solution: Set y = to obtain the ^-intercepts, which are ± 3; and set x = to obtain the ?/-intercepts, which are ± 2. Solve for y and obtain y = ± q V9 - x 2 . Construct a table and draw the curve, as shown in Fig. 12-5. This illustrates an ellipse. X y -3 -2 ±1.49 - 1 ±1.89 ±2 1 ±1.89 2 ±1.49 3 +x Fig. 12-5. Note that the numerical value of x must be equal to or less than 3 in order to yield real values of y. 224 Equations in Quadratic Form Sec. 12-10 Example 12-26. Graph 4x 2 - 9y 2 = 36. Solution: Setting y = 0, we find that the x-intercepts are ± 3. Setting x = 0, however, results in the equation y 2 = - 4. Hence the curve has no ^/-intercepts. Solving the given equation for y } we have 2 y = ± 3 v^ 2 9. The accompanying table is constructed. O X y -6 ±3.5 -5 ±2.7 -4 ± 1.8 -3 -2 imaginary 2 imaginary 3 4 ±1.8 5 ±2.7 6 ±3.5 Fig. 12-6. Note that in this case the numerical value of x must be equal to or greater than 3 in order to give real values of y. The graph of the given equation is shown in Fig. 12-6. This illustrates a hyperbola. EXERCISE 12-8 Identify and graph each of the following. 1. x 2 + y 2 = 25. 2. 4x 2 + 9x 2 = 36. 3. 4x 2 - 9y 2 = 0. 4. 4x 2 - 9y 2 = 36. 5. x 2 - 4*/ 2 = 16. 6. x 2 + 9y 2 = 0. 7. y = x 2 - 3x + 2. 8. x?/ = - 4. 9. 5a; 2 + 9xy = 28?/ 2 . 10. 2x 2 + y 2 - 4y =4. 11. 5xy = 2x + y. 12. 3x 2 - 4x?/ + 2?/ 2 - 6x + 3?/ = 7. 13. x 2 + x# - 2y 2 + 3y - 1 = 0. 14. 4a- 2 - 4xy -f Z/ 2 - 2x + 4?/ - 12 = 0. 15. xy + y 2 - y - 2x - 2 = 0. 16. x 2 - 3x - 3?/ 2 + 18y = 27. 17. 2x 2 - xy - 28# 2 = 0. 18. 4x 2 + \.xy - 3?/ 2 + 4x + 10t/ = 3. 19. 9x 2 _ 24xy + 16^/2 + 3x - 4?/ = 6. 20. 4x 2 + Zxy + \.y 2 - Sx - %y = 24. 12-11. GRAPHICAL SOLUTIONS OF SYSTEMS OF EQUATIONS INVOLVING QUADRATICS In Chapter 9 we solved systems of two or more linear equations both algebraically and graphically. Frequently, however, simul- taneous systems include one or more equations of the second or higher degree. We have, therefore, to consider the problem of find- ing systems of values of the unknowns x and y that satisfy two equa- Sec. 12-11 Equations in Quadratic Form 225 tions, one of which is quadratic and the other of which is linear or quadratic. We shall begin by illustrating some graphical solutions of several types of systems. The graphical method yields only the real solu- tions of a system, but it may prove advantageous in suggesting solutions and interpreting results. In general, this method yields at best only approximate solutions. The graphs should be drawn as accurately as possible. Example 12-27. Solve graphically the system x 2 - Sx + 3y = 0, * x - 3y + 6 = 0. Solution: Solving each equation for y in terms of x f we have 8x - x 2 , x + 6 y = — 3 — and y = — y- • Construct tables of values, and draw both graphs, using the same coordinate system, as shown in Fig. 12-7. 8x - x 2 X y - 3 - 1 -3 1 7/3 2 4 3 5 4 16/3 5 5 6 4 7 7/3 8 9 -3 6 y =■ x + 6 +x (A): .r2-.8.l: + 3 l r«0 , (*>.r-.3/+6-0 Fig. 12-7. The line and the parabola are seen to intersect at the points (1, 7/3) and (6, 4). It follows that these points represent common real solutions, possibly only approxi- mate, of the system. A check by substitution shows that the real solutions are, in fact, x = 1, y = 7/3; and x = 6, y = 4. The solution of Example 12-27 suggests a procedure for finding graphical solutions of quadratic equations in one unknown. 226 Equations in Quadratic Form Sec. 12-11 Example 12-28. Solve the equation x* - x — 2 = graphically. Solution: Since x 2 = x + 2, both sides of this equation may be set equal to y. Thus, the original equation is replaced by the system J y = z 2 , I ?/ = x + 2. a; 2/ = x* ±1 1 ±2 4 ±3 9 ±4 16 X y =x +2 -2 2 From the accompanying tables of values of x and y, the graphs shown in Fig. 12-8 are constructed. The graphs show that the line and the parabola intersect at the points (— 1, 1) and (2, 4). Since these points are common to both graphs, their abscissas must satisfy the equation x 2 - x + 2. Hence, the required roots of the original equation x 2 — x — 2 = arc x = — 1 and x = 2. *** Fig. 12-8. (A): j? 2 +4^- 2 -16 ^; 3.r2_2 r 2 «6 Fig. 12-9. Example 12-29. Solve graphically the system x 2 + 4y 2 = 16, 3x 2 - 2?/ 2 = 6. Solution: From the first given equation, 4?/2 = 16 - x 2 , and From the second equation, and V = =fc ^ V16 - z 2 . 2*/ 2 = 3z 2 - 6, Sx 2 -6 Sec. 12-12 Equations in Quadratic Form 227 The necessary tables are given here, and the graphs are shown in Fig. 12-9. I X y = =b ^ V16 - z 2 ±2 1 ±2 Vl5 2 db V3 3 4V7 4 - 1 ±iVl5 -2 ± a/3 -3 = ±V7 -4 X , i /3^2 - 6 imaginary ±1 imaginary =fc v/2 2 ±V3 3 ± A/T0i5 4 =b V21 -2 ± \/3 -3 ± V105 -4 ±\/21 The ellipse (A) and the hyperbola (B) are seen to intersect in the following four distinct points: (2, V3), (2, - V3), ( - 2, V3), ( - 2, - V3). These values of x and ?/ already appear in the tables used for constructing the graphs, and need not be checked by substitution in the original equations. However, such checking is usually desirable. EXERCISE 12-9 Solve each of the following systems of equations graphically. 1. (x - 2y +3=0, 2. U +?/ =4, 3. (x - y + 1 = 0, U 2 = 3y. 1 ?/ 2 = 2s. 1*2 + ? y2 = 25. 4. L> = 3x, 5. |* + 2/ = 6, 6. hx + 22/ = 6, Ux + y = 6. 1 .r 2 = ?/. \ X y = - 12. 7. |2.r -2/ =4, 8. fj* - y* = 16, 9. hx + 32/ = 25, Ixj/ = 6. 1 x + 32/ = 4. U*2 + ^2 = 25. 10. Lr* - 2v/ 2 - 4 = 0, 11. | x 2 + 7/2 = 13, 12. |9z 2 +42/2 =36, 1*2 _ (ty = o. 1 x 2 = 122/. 1*2 + y 2 = si. 13. jO-r 2 + 25s/ 2 =225, 14. L' 2 +?/ 2 = 20, 15. |*2 -f- j/a = 20, }z 2 +t/2=4. 16. \y* -x 2 = 12. | X + 7/ = 0, {*2 + ^2 = 8. [4x 2 + 92/ 2 = 100 12-12. ALGEBRAIC SOLUTIONS OF SYSTEMS INVOLVING QUADRATICS As in the case with linear equations discussed in Section 9-1, it may happen that in a system of equations involving quadratics 228 Equations in Quadratic Form Sec. 12-12 part of the graph of one equation coincides with part of the graph of the other. Such a condition gives rise to infinitely many solu- tions. Usually, however, there are only a finite number of points of intersection of the graphs corresponding to the given equations, and the algebraic problem consists of finding the pairs of numbers (x, y) which satisfy both equations. We can say in this case that two simultaneous equations in x and y, of degrees m and n, respec- tively, can have at most mn solutions. Thus, a system of one linear and one quadratic equation can have at most two solutions, and a system of two quadratics can have at most four solutions. When a system consists of two quadratic equations, the algebraic solution usually leads to a fourth-degree equation in one of the unknowns. Since we have not presented a general method of solving a fourth-degree equation, we shall consider here only systems whose solutions can be effected by the theory of quadratic equations. The methods of procedure in some of the more important types are shown in the following three cases : Case 1. One Linear and One Quadratic Equation. A system of this type can always be solved by the method of elimination by substitution. Example 12-30. Solve the system x - 3?/ + 6 = 0, x 2 - Sx -f 3y = 0. Solution: Solve the linear equation for y in terms of x, obtaining x + 6 »=-3— Substitution for y in the quadratic equation yields x* - 8x + 3 p-jp*) = 0. Collecting terms gives x 2 - 7x + 6 = 0. The roots of this equation are x = 1 and x = 6. Substituting these values in the linear equation, we obtain y = 7/3 and y = 4. Hence, the solutions are x = 1, y = 7/3; and x = 6, y = 4. These values can readily be verified as solutions of the given system. Note that they correspond to the coordinates of the points of intersection in Fig. 12-7. Example 12-31. Solve the system x + 2y + 4 = 0, x 2 + 4y* - 2x - 3 = 0. Sec. 12-12 Equations in Quadratic Form 229 Solution: Solve the linear equation for 2y y to obtain 2y = - (z + 4). Substitute in the quadratic and collect terms. We then have ' x 2 + (a + 4) 2 - 2x - 3 = 0, or 2x 2 + 6x + 13 = 0. One solution is - 3 + t V17 5 + t V17 * = 2 ' V= 4 The other solution is _ - 3 - i -v/17 5 - f V17 *= 2 ' ^ = 4 Since these values are imaginary, the graphs of the two given equations do not intersect. Case 2. Two Equations of the Form ax 2 + by 2 = c. When the system consists of two equations containing only squared terms in each unknown, it can be solved for x 2 and y 2 by the methods used for linear systems in Section 9-2. Example 12-32. Solve the system x 2 + 2y 2 = 17, 2 X 2 _ y 2 = 14. Solution: To eliminate y 2 , multiply the second equation by 2 and add the two equations, as follows: x 2 + 2y 2 = 17 4x 2 - 2y 2 = 28 5.r 2 = 45. Solving for x, we have x = ±3. Now substitute 9 for x 2 in the first of the original equations. Then 2y 2 = 17 - 9 = 8, or y = ±2. Hence, we have the following four solutions: (3,2), (3,-2), (-3,2), (-3,-2). These may be written (3, =b 2), ( - 3, d= 2). Case 3. Two Equations of the Form ax 2 + bxy + cy 2 = d. If the system is of this type, the solution is effected by elimination of the constant term. The resulting equation is then solved for one unknown in terms of the other. This procedure gives us two linear equations in x and y which may be combined with either of the given quadratic equations to form two systems of the type con- sidered in case 1. 230 Equations in Quadratic Form Sec. 12-12 Example 12-33. Solve the system x 2 - xy + 2y 2 = 1, 2x 2 - 2xy + Sy 2 = 3. Solution: Multiply the first equation by 3 and subtract the second given equation from the new equation, as follows: Sx 2 - Sxy + 62/ 2 = 3 2x 2 - 2xy + Sy 2 = 3 x 2 - xy - 2y 2 = 0. Factor, to obtain (x + y) (x - 2y) = 0. Hence, x + y = or x — 2i/ = 0. We may combine each of these two equations with the first given equation to form the following two systems: f x 2 - xy + 2y 2 = 1, I x + y = 0; and fa; 2 -2;i/+ 2?/ 2 = 1, \ a; - 2y = 0. We then proceed by the method for case 1. The solutions of the given system consist of the solutions of these two systems. Hence, we have x = 1/2, y = - 1/2; x = - 1/2, y = 1/2; x = 1, y = 1/2; * = - 1, ?/ = - 1/2. Occasionally, another method is effective in connection with systems described under case 3. The following example illustrates this method. f & + ®y* = 37, Example 12-34. Solve the system xy = 2. Solution: Multiply the second equation by 6, to obtain §xy = 12. Add this to the first equation, to obtain x 2 + 6xy + 9ij 2 = 49. Also subtract Qxy = 12 from the first equation to obtain x 2 - Qxy + $y 2 = 25. The left side of each of these new equations is a perfect square. We chose the multiplier of xy } which is 6 in this case, so as to obtain perfect squares. We now have the system / (x + Sy) 2 = 49, \ (x - Sy) 2 = 25. Hence, x + Sy = 7 or x -f Sy = - 7. Also, x - Sy = 5 or x - Sy = - 5. We now solve the following four systems of linear equations: Iz + Sy = 7, ix+Sy= 7, fa + 3y = - 7, f * + Sy = - 7, \x-Sy = 5; \x - 3y = - 5; \z-3y=5; \s - 3y = - 5. Sec. 12-13 Equations in Quadratic Form 231 An equation in x and y is said to be symmetric in x and y if the equation is unchanged when x and y are interchanged. When a system consists of two quadratic equations both of which are sym- metric in x and y, the substitutions x = u + v, y — u — v will give an equivalent system which may in some cases be solved by previ- ous methods. EXERCISE 12-10 In each of the problems from 1 to 21, solve the given system of equations algebraically. 1. f x - 2y + 3 = 0, 2. f x + y = 4, 3. f x - y + 1 = 0, \x 2 = 3y. \y 2 = 2x. \x 2 + y 2 = 25. 4. Uj 2 = 3a;, 5. 13a; + 2y = 6, 6. 1 2x - y = 4, 1 3a: + y = 6. \xy + 12 = 0. \xy = 6. 7. jx 2 - ?/ 2 = 16, 8. j 8s + 3y = 25, 9. jy - x = 2, jx + 3y = 4. 1 4a; 2 + 2/ 2 = 25. 1 x 2 + y 2 - 2x - 4y = 20. 10. Is 2 - 2y 2 = 4, 11. \x 2 + y 2 - 10, 12. Uz - 3) 2 + (y - l) 2 = 16, \x 2 -9y=0. I a; 2 =9i/. I (x - l) 2 + (?/ - l) 2 = 12. 13. J 9a; 2 + Ay 2 = 36, 14. \dx 2 + 25y 2 = 225, 15. \x 2 + y 2 = 20, } x 2 + 2/ 2 = 81. \ x 2 + 2/ 2 = 4. J?/ 2 - a; 2 = 12. 16. (xy = 2, 17. 1 1 _ 3 18. (x 2 + 3xy = 10, \ X 2 - V 2= 3. a; 2/~2' 1^2/ = 3. la; + 2/ = 3. 19. fa; 2 + Sxy = 28, 20. f 3a; 2 + ?/* = 28, 21. fa;?/ + ?/ 2 = 12, [xy + 4?/ 2 = 8. [4a; 2 - xy + y 2 = 40. jxy = 2a; 2 - 24. 22. Complete the solution of the system in Example 12-34. In each of the problems from 23 to 26, solve the given system by the method of Example 12-34. 23. fa; 2 + y 2 = 50, \xy = 25. 25. \x 2 + y 2 = 25, (xy = 12. Solve each of the following symmetric systems. 27. fa; 2 + y 2 = 4, jxy + a; + y = 10. 29. frca/y + y 2/ x = 5^ \x + 2/ = 2. 12-13. EXPONENTIAL AND LOGARITHMIC EQUATIONS An equation in which the unknown occurs in an exponent is called an exponential equation. Such an equation is usually solved by taking the logarithm of each side and solving the resulting equation. When this latter equation is in linear or quadratic form, it may be solved by preceding methods. 24. fa; 2 + 4y 2 = 13, \xy = 3. 26. fa; 2 + y 2 = 144, \xy = 56. 28. fa; 2 + y 2 - x - y = 2, \xy + 3a; + 3y = 2. 30. |x 2 + i/ 2 = 13, \3x 2 + 2xy + 32/ 2 = 42. 232 Equations in Quadratic Form Sec. 12-13 Example 12-35. Solve for x: 5* +3 = 625. Solution: Write the equation in the form 5*+ 3 = 5 4 . This equation is satisfied if and only if x + 3 = 4, that is, x = 1. Example 12-36. Solve the equation 2 3 * +1 = 3 4 *. Solution: Taking the logarithm of each side to the base 10, we get log (2 3 * +1 ) = log (3 4 *), or (Sx + 1) log 2 = 4z log 3. Therefore, log 2 4 log 3 - 3 log 2 From Table III, log 2 = 0.3010 and log 3 = 0.4771. Substituting these values, we have 0.3010 X ~ 4 • 0.4771 - 3 • 0.3010 ' or O3010 X ~ 1.0054 ~ °- 2 " 4 ' Example 12-37. Solve for x: log (x + 1) = 1 - log (3s + 2). Solution: Collecting terms containing logarithms on one side and writing that member as a single logarithm, we have log (* + 1) + log (3x + 2) = 1, or log (x + 1) (3s + 2) = 1. Writing the members in exponential form, we have (x + 1) (3* + 2) = 10* = 10. This equation reduces to 3x 2 + 5x - 8 = 0. § Solving for x, we find that x = lorx= ~^* o g Checking, we find that x = 1 satisfies the original equation, whereas x = — - o gives rise to logarithms of negative numbers. Since negative numbers do not have real logarithms, this latter value of x is not to be used. Example 12-38. Solve y = log, (x + y/l + z 2 ) for x in terms of y. Solution: Write the given equation in exponential form, obtaining & = x + \/l + x 2 . Transpose and square to remove the radical. The result is e 2 * - 2x& -1=0. Solving this equation for x, we have e*v - 1 Sec. 12-14 Equations in Quadratic Form EXERCISE 12-11 Solve each of the following equations for the unknown x. 1. 2* = 64. 4. 10* = 0.0001. 7. 5* = 15. 10. (3*) (2*) = 36. 13. log* 2 = 0.6932. 16. 5.03 = (S.Viyii*- 1 ) 19. log (3x - 5) = 3 - log 7, 21. x l0 « ( * 3) = 1000. 23. e 2 * = 4.83. 26. e* 2+2 *- 2 = 16. e x + e~* 2. 4*+* = 256. 5. 4* = 24. 8. 3* = 17. 11. 5"* = 2.403. 14. s 3 « l4 « =0.04681. 17. 3* +2 = 2(5*). 233 3. 3 3 * +l = 243. 6. 2 4 * = 2 3 *-*. 9. 3(2*) = 6*. 12. log, 8 = 0.4136. 15. (1.5)* = 32. 18. log 2 x = 3. 29. y = 24. c ( * 2) = 9.436. 27. ac + 6c* +1 = 1. e* + e~* 30. y = 20. log (4x - 1) = 1 - log (6x + 2). 22. log 2 (x - 1) + log 2 (x + 3) = 3. 25. 4e*+* = 7. 31. e 4 * - e 2 * - 10 = 0. e* - e-* 32. c* + 4e~* = 5. 12-14. GRAPHS OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS The graph ofy = log x is shown in Fig. 12-10. By assigning values to x, one finds corresponding values of y from Table III. A few pairs of values are shown in the accompanying tabulation. X y 0.1 - 1 0.2 -0.70 0.3 -0.52 0.5 -0.30 1 2 0.30 3 0.48 4 0.60 10 1. +x Fig. 12-10. The graph of y = e* may be obtained from a table of exponential functions. Here, however, we shall proceed as follows: Take the logarithm of each side to the base 10, to obtain log y = x log e. 234 Equations in Quadratic Form Sec. 12-14 Prepare the accompanying table, and construct the graph in Fig. 12-11. *►* X loge x log e y 0.434 1.0 1 0.434 0.434 2.7 2 0.434 0.868 7.4 3 0.434 1.302 20.0 4 0.434 1.736 54.5 • Graph each of the following. 1. y = log 7 x. TT- x i l°gio x mnt:log7X =__. 4. y = 10-* 2 . 7. y=3+4(20. 10. 2/ = 100e - 06 *. EXERCISE 12-12 2. x = log 7 ?/. 5. 7/ = 02. 8. 2/ = e~*. 11. y = 7 3 *- 1 . 3. ?/ = 3-. 6. 2/ = 16(5-). 9. y = log c x. 12. y = 3.9 2 *- 3 . 13 Theory of Equations 13-1. INTRODUCTORY REMARKS With the ever-increasing importance of mathematics in engi- neering and the physical sciences, problems are constantly occur- ring that involve the solution of equations. Often these equations are of the simple algebraic or trigonometric types which we have already learned to solve. There are many other problems, however, which require the solution of equations of higher degree than the second and of some types of somewhat more complicated tran- scendental equations. Equations of the third and fourth degree can be solved by methods analogous to those which we used for quad- ratic equations. Because of their complexity, however, these methods are seldom used. It has been proved that no such proce- dures exist for equations of degree higher than the fourth. In this chapter we shall consider various properties of poly- nomial equations in general. Some of these properties will be of considerable use in later studies of mathematics, while others are considered here merely for the aid they give us in determining roots of equations. 13-2. SYNTHETIC DIVISION A simplification of the ordinary method of long division, called synthetic division, will be presented here. This abbreviated method not only enables us to quickly find the quotient and remainder when a polynomial f(x) = a x n + a x x n - 1 H V a n is divided by a binomial of the form x — r, but also affords a simple process for substituting values of the variable into a polynomial. We shall divide 2x s — 9x 2 + 13x + 5 by x — 3 to illustrate the pro- cedure in synthetic division as compared with that of long division considered in Section 1-19. 235 236 Theory of Equations Sec. 13-2 By long division we have : 2x 3 - 9a; 2 + 13s + 2a; 3 - 6a; 2 - 3a; 2 + 13x -3a; 2 + 9a; 4a; + 4a; — 5 12 x - 3 2s 2 - 3x + 4 + 17 Thus, the quotient is 2x 2 — Sx + 4, and the remainder is 17. Since like powers of x are written in the same vertical column, the work may be shortened by writing only the coefficients, as in the following schematic arrangement: 2 -9 -6 -3 + 13 + 13 + 5 1 -3 2 2 -3 + 4 -3 + 9 + 4 + 4 + 5 - 12 + 17 Next, we note that the first term in the divisor x — r need not be written, since the divisor is always linear, and the coefficient of x in it is always unity. Moreover, it is not necessary to write the first term in each row that is to be subtracted, since its coefficient is always the same as that of the term directly above. Also, only the first term of each partial remainder needs to be written down, for the second term is the same as the term directly above it in the first row. Finally, the coefficients in the quotient need not be written, since these are precisely the leading coefficient in the dividend and the remaining partial remainders, excepting the last. Hence, we may indicate the process in the following way : 2-9+13+5 1 -3 - 6 -3 + 9 - 12 + 17 Sec. 13-2 Theory of Equations 237 This scheme can be written compactly as follows : 2 - 9 13 5 | - 3 - 6 9 12 2-3 4 17 If we replace —3 by +3 in the divisor and add the partial prod- ucts in the second row instead of subtracting them, we obtain the same result. The synthetic division then takes the following form : 2-9 13 5 [_3_ 6 -9 12 2-3 4 17 Here the numbers 2, —3, and 4 in the third row are the coefficients of the quotient, and the last number 17 is the remainder. We can now outline the procedure for synthetic division. Note that in every step of the procedure, immediate reference is made to the illustrative example. To divide f(x) — a x n + a x x n ~ x H h a n by x — r, first arrange f(x) in descending powers of x, writing zero for the coefficient of any missing power of x. Then arrange the numbers involved in the process in three rows, as shown in the following steps : Step 1. In the first row, write the coefficients in f(x) in order, as a , a U " a n . At the right, put the constant term of the divisor with its sign changed. We have then Example 0o ai c&2 • • • a„ | r_ 2 — 9 13 5 | 3_ Step 2. Bring down the first coefficient a of f(x) into the first place of the third row. Thus, we have Example ao a\ a,2 • • • a n \_r_ / 2 — 9 13 5 [_3_ ( *a *2 Step 3, Multiply a by r, and write the product a r in the second, row under a x . Bring down the sum of a x and a r into the third row. Thus, we now have Example oo ai fl2 • •• a n |_r_ 2 — 9 13 5 1 3 apr 6 . ao (<W + a>i) 2-3 238 Theory of Equations Sec. 13-2 In the example, multiply 2 by 3 and write the product 6 in the second row below —9. Then add —9 and 6, writing the sum —3 in the third row. Step 4. Multiply a r + a x by r, place the product in the second row under a 2 , and add. Continue this process until finally a product has been added to a n . The complete solution of the illustrative example follows : 2-9 13 5 1 3 6-9 12 2-3 4 17 In the last operations, -3 is multiplied by 3, and the product -9 is written below 13. The sum of 13 and —9 equals 4. Finally the product of 4 and 3, or 12, is written below 5 and added to a n = 5 to give 17. Step 5. When the process is completed, the last number in the third row directly below a n is the remainder. The other numbers in this row, read from left to right, are the coefficients of powers of x in the quotient arranged in descending order. The entire synthetic process of dividing a polynomial f(x) by x — r, although it is somewhat complex notationally, can be con- veniently exhibited as follows : ao ai a,2 • • • a n _i a n | r bpr bir • • • b n _ 2 r b n -\r bo b\ 62 • • • &, t -i It Here the expressions for the coefficients b , b u b 2 , ■ • • , b n 1 of the powers of x in the quotient are 6 = a , &i = a„ r + a l9 b< 2 — a r 2 + ai r + a 2 • • • , 6 n -i = do ^ n " 1 + Q>i r* 1 ' 2 H + a n -i. Hence, the quotient may be written (13-1) q (x) = b x*' 1 + 61 x n ~ 2 + ••• + & n -i. Also, the remainder assumes the form (13-2) R = a r n + air"- 1 H h a n ^r + a n . The expression for R is precisely the result of substituting r for x inf(x). In other words (13-3) «=/(r). Finally, if we subtract the product of a? — r and <?(#) from /(#), we obtain the remainder R = f(r). Or, if we transpose the product, we have the usual statement found in discussions of division. That is, (13-4) /(*) = (*-r).a(aO+/(r). Sec. 13-2 Theory of Equations 239 The following examples will further illustrate the process of synthetic division. Example 13-1. Divide 3x 4 - 4a; 2 + x - 2 by z + 2. Solution: Sincere - r = z -f 2, we haver = - 2. Writing zero for the coefficient of the missing power x 3 , we have the following result: 3 0-4 1-2 1-2 - 6 12-16 30 3-6 8-15 28 Note that the first coefficient 3 is brought down into the first place of the third row. Next 3 is multiplied by — 2, and the product, which is — 6, is written in the second row under 0. The sum of and - 6, or - 6, is written in the third row directly belowO. Proceeding in this way, we find that the quotient is 3a? 3 — 6a; 2 + 8a; — 15 and the remainder is 28. Example 13-2. Given /(a;) = a; 3 - 2x 2 + 5s - 4, find a)/(-D;&)/(D;c)/(3). Solution: By (13-3), f(r) equals the remainder obtained in the division of f(x) by a; — r. Hence, we have the following results : 1-2 5-41-1 a) - 12. - 1 3 - 8 Therefore, /( - 1) = - b) 1 1 -3 -2 1 8 5 - 1 - 12 - 4 u 4 Therefore, /(l) = 0. c) 1 1 - 1 -2 3 4 5 3 - 4 ul 24 Therefore, /(3) = 20. 1 1 8 20 EXERCISE 13-1 In each of the problems from 1 to 15, divide the first function by the second, to find the quotient and the remainder, by using synthetic division. 1. x 2 - Sx + 7, x - 1. 2. a: 3 - x 2 - Sx + 6, x - 3. 3. x 2 - 5x + 6, x - 4. , 4. a; 3 - 3a; 2 + 6a: - 6, x - 3. 5. a; 4 - 3a; 2 - 6x + 6, i + 3. 6. a; 3 - 3x 2 + 6.c - 24, a - 4. 7. 2a; 4 - a; 3 - 6a; 2 + 4a; - 8, a; - 2. 8. 2a; 3 + 3a; 2 + 4, a; -f 2. 9. .c 3 + 3a; 2 - 2a; - 5, x - 2. 10. 2a; 5 - 3a; 3 + 2a; + 1, x + 2. 11. a; 4 + x 3 - 59s 2 - 69a; + 030, z 2 - x - 42. (Hint: Factor, the divisor and divide successively by each factor.) 240 Theory of Equations Sec. 13-2 12. x* - 3a? 3 + 3a; 2 - 3x + 2, x 2 - 3a; + 2. 13. x 5 + 2a; 4 - 21a; + 18, x 2 + 2x - 3. 14. a w — 1, a — 1. 15. x n — i/", x — ?/. For each of the following polynomial functions, find the indicated values by the method of synthetic division : 16. f(x) = 3a; 3 - 7x 2 - 5s + 6. Find/(1) and/(- 4). 17. f(x) = a; 4 - 2a; 3 + 2a; 2 - 5x + 2. Find/( - 2) and/(0). 18. fix) = x 4 - 4a; 3 - 4a; 2 + 24a; - 9. Find/(3) and/(9). 19. fix) = a; 4 - 3a; 3 - 13x 2 + 21x + 18. Find/(1) and/(3). 20. f{x) = 7a; 4 + 37a; 3 - a; 2 - 14a; + 4. Find/(1) and/(- 5). 13-3. THE REMAINDER THEOREM In Section 13-2, it was shown that the remainder in the division of a polynomial by a binomial x — r can be found without actually performing the division. Thus, in establishing (13-3), we have proved the following theorem. Remainder Theorem. If a polynomial fix) is divided by x — r, the remainder is the value of f(x) for x = r; that is, the remainder is/(r). It follows from this theorem that f(x) is exactly divisible by x — r if and only if f(r) = 0. Hence, we have proved also the fol- lowing theorem. Factor Theorem. If f(r) = 0, then x — r is a factor of the poly- nomial /(#), and conversely. Example 13-3. Is x + 3 a factor of a; 4 - 2a; 3 + 3a; 2 - 5? Solution: Here f(x) = a; 4 — 2a; 3 + 3a; 2 — 5. Also, x — r = a; -f 3, and r = — 3. According to the factor theorem, /( — 3) must equal zero if a; -f 3 is to be a factor of f(x). But /(-3) = (-3) 4 -2 (-3) 3 +3(-3) 2 -5 =81 +54 +27 -5 = 157. Therefore, since /( — 3) ^ 0, x -f 3 cannot be a factor of a; 4 - 2a; 3 + 3a; 2 — 5. Example 13-4. Given fix) = a; 3 - 3a; 2 + 5a; - 6. Show that /(2) = and, therefore, that x — 2 is a factor of fix). Solution: Since /(x) = x 3 - 3x 2 + 5a; - 6, we have by substitution /(2) = (2) 3 - 3 (2) 2 + 5 (2) - 6 = 0. From the fact that fix) equals zero when x = 2, it follows from the factor theorem that x — 2 is a factor of fix). The student should check this result by synthetic division and find that fix) = a; 3 - 3a; 2 + 5a; - 6 = (a; - 2) (a; 2 - x + 3). Sec. 13-4 Theory of Equations 241 Example 13-5. Find under what condition (x -f a) is a factor of x* + a n , where n is an integer and a j* 0. Solution: In this case, /(#) = r» + a n , and /( — a) = ( — a) w -f o n . This sum can equal zero only if ( — a) n = — a n , that is, only when n is odd. EXERCISE 13-2 In each of the problems from 1 to IS, determine if the second function is a factor of the first. If it is a factor, find another factor. 1. 2x* - Gx 2 + x + 6, x - 2. 2. x* + 4a; 2 + 5z + 6. x 2 + a + 2. 3. z 3 + 2z 2 - 3x - 1, x - 1. 4. x 7 - 1, x - 1. 5. a: 3 - x 2 - \\x + 15, x - 3. 6. z fi + 243, x + 3. 7. x 8 - 256?/ 16 , x - 2?y 2 . 8. 2a; 3 -f 3x* - 9z - 111, a - 1. 9. z 4 - 4z 3 - x 2 + 16a; - 12, :r + 1. 10. x 5 + 5a; 4 + 20a; 3 + 60a; 2 + 120z + 120, x + 1. 11. 5a; 3 4- ah (5 - 6«6 2 )a; a - 2a 2 6 2 (5 + 3a6 2 )a; + 12a 4 6 6 , & - aft. 12. 2x l - 3a; 3 - 3a; - 2, & + 2. 13. a; 5 - 10a; 4 + ISa; 3 - 24.1' + 75, x - 2. 14. 2a; 4 - 31a; 3 + 21a; 2 - 17a; + 10, a; + 1. 15. 12.x 4 - 40a; 3 - x 2 + Ilia; - 90, 2a; - 3. 16. 12a; 3 - 22j 2 - 34a; + GO, 3a; + 5. 17. 24a; 4 - 122a; 3 + 159a; 2 - 7a; - 00, 3a; - 4. 18. 24x* - 74a; 4 - 85x« + 311a; 2 - 74a; - 120, 2a; + 1. 19. Show that the equation a; 4 + 2x B - 9r 2 - 2x +8=0 has the roots 1, 2, - 1, - 4. 20. Find all roots of the equation 12a; 4 - 40a' 3 - x 2 + Ilia; - 90 = 0, given that 3/2 is a double root, that is, that (a- — 3/2) 2 is a factor of the left side. 21. Prove that a — b is a divisor of a n — b n for every positive integral value of n. 22. Prove that a + b is a divisor of a n — b n if n ir? a positive even integer. 23. Prove that a + b is a divisor of a n + b n if n is a positive odd integer. 24. Prove that neither a + b nor a — b is a divisor of a n -f b n if n is a positive even integer. 13-4. THE FUNDAMENTAL THEOREM OF ALGEBRA We usually assume that every algebraic equation with real or complex coefficients has at least one real or complex root. Although the existence of such a root is not to be taken for granted, a proof is beyond the scope of this book. Accordingly, we shall accept the following theorem as true. Fundamental Theorem of Algebra. Let f(x) be a polynomial of degree n with complex coefficients. Then the algebraic equation f(x) =0 has at least one complex root. The first complete and rigorous proof of the fundamental theorem was given by Gauss in the beginning of the nineteenth century. Since that time, many proofs have appeared, but most require knowledge of the theory of complex functions. 242 Theory of Equations Sec. 13-4 By a repeated application of the fundamental theorem, it can be shown that the number of roots of any polynomial equation with real or complex coefficients is equal to the degree of the poly- nomial. We shall state the following theorem and give its proof. Theorem. If f(x) is a polynomial of degree n, the equation f(x) = has exactly n roots. Proof. Let f(x) be a polynomial of degree n with real or com- plex coefficients. By the fundamental theorem, there is a number r x such that /(ri) =0. Then, by the factor theorem, /Or) = (x - n) • qi(x), where qi(x) is a polynomial of degree n — 1 whose leading coeffi- cient is a . Likewise, qi(x) = has a root by the fundamental theorem. If we denote this root by r 2 , then q x (r 2 ) = 0. Also, qi(x) = (x - r 2 ) • ?2(s), where <z 2 (#) is of degree n — 2 with leading coefficient a . Similarly, # 2 (x) = also has a root. We can continue in this way until we come to a polynomial of the first degree with root r n . Then (13-5) fix) = a (x - n) (x - r 2 ) • • • (x - r n ), where a is the coefficient of # in our final first-degree polynomial. Since f(x) equals zero when we substitute for x any one of the n numbers r u r 2 , • • • , r n , it follows that the equation f(x) = has at least the roots r u r 2 , • • • , r w . Moreover, there are no other roots. For, suppose that r is some root other than r u • • • , r n . Substitution of r for x in (13-5) yields f(r) = a (r - ^i) (r — r 2 )"*(r — r n ). The right side of this equation cannot equal zero, since no one of the factors can equal zero. Therefore, f(r) ^0, and f(x) =0 has no more than the n roots found before. It may happen that a certain root appears more than once among the numbers r lf r 2 ,'> ?V In that case it will be counted as many times as the corresponding equal factors appear in (13-5). If a certain factor (x — r) appears m times in (13-5), then r is said to be a root of multiplicity m. A root is called a simple root, a double root, a triple root, and so on, the proper name depending on how many times the same factor appears. Hence, combining the state- ments that "f(x) = has at least n roots" and "f(x) = has no more than n roots," we can conclude that a polynomial equation of the nth degree has exactly n roots, a root of multiplicity m being counted as m roots. Sec. 13-5 Theory of Equations 243 It is to be noted that a rigorous proof of this theorem requires the use of induction. (See Chapter 16.) 13-5. PAIRS OF COMPLEX ROOTS OF AN EQUATION We wish to remind the student that while an equation of the nth degree has n complex roots, the number of real roots may be less than the degree of the equation. For example, x 2 + 1 = has no real roots. Determination of the number of real roots may be sim- plified by use of the following theorem. Theorem. If all the coefficients of f(x) — are real numbers, and if the complex number a + bi is a root of f(x) — 0, then the conju- gate a — bi is also a root. It is understood that a and b are real and b¥=0. Proof. Let x in f(x) — a x n + a x x 71 ' 1 H \- a n be replaced by a + bi. Then we have f(a + bi) = a (a + bi) n + a x {a + bi) 71 - 1 H h a n . If we expand the powers of a + bi by the binomial theorem and simplify the resulting expression, then all terms which contain even powers of i will be real, while all terms which contain odd powers of i will be pure imaginary. Denote by P the aggregate real part, and by Q the aggregate imaginary part. Then we have f(a + bi) = P + Qi = 0. Hence, in accordance with (11-4), P = Q = 0. Now, replace x by a — bi in f(x) - 0. In those terms of f(a — bi) in which — bi is raised to an even power, the result will remain the same as in f(a + bi) . However, all terms in f(a — bi) in which —bi is raised to an odd power will have their signs changed. Hence, /(a - bi) = P - Qi. But, since we have shown that P — Q = 0, we conclude that P - Qi = 0. In other words, a — £n is also a root of f(x) = 0. Example 13-6. Solve x A - z 3 - 2x 2 -{- 6# - 4 = 0, one root being 1 + i. ; Solution: According to the last theorem, both 1 + i and 1 — i are roots of the given equation. Using Eqs. (12-3) and (12-4) for the sum and product of two roots, we find that these conjugate complex numbers are roots of a; 2 — 2x +2 =0. Dividing the original polynomial by this quadratic function, we get the quotient x 2 + x — 2. Solution of the equation x 2 -j- x — 2 = yields the remaining desired roots, x = — 2 and x = 1. 244 Theory of Equations Sec. 13-5 EXERCISE 13-3 1. Solve x 8 + x 2 . - 2x -f 12 = 0, one root being 1 + \/3i. 2. Solve a: 3 + 3x 2 + 12x - 16, one root being -2-2 V" 17 ^ 3. Solve z 4 - 2x3 _ j X 2 + 1&. - 18 = 0, one root being 1 - i. 4. Solve x 4 + 3.t 3 -f 7z 2 + 6z + 4 = 0, one root being - 1 - s/Si. 5. Solve x* - 8z 4 + 27z 3 - 46x 2 + 38a; - 12 = 0, one root being 2 - V2i, and one root being 1. 6. Solve # 4 + x 2 + 1 = by considering the equation to be a quadratic equation in x 2 . 7. Find a real cubic equation, two of whose roots are 2 and 1 -f 2i. 8. Find a real equation of lowest degree having the roots i and 1 — i. 13-6. THE GRAPH OF A POLYNOMIAL FOR LARGE VALUES OF x. In graphing a polynomial function, it is helpful to know the location of points on the curve for numerically large values of x. It can be shown that, when x is numerically sufficiently large, the term a x n of highest degree is numerically larger than the sum of all the other terms combined. Therefore, the sign of this term determines the sign of the entire polynomial. Let us consider the values of the function f{x) = x' 6 + 5x 2 - Ix — 13 as x assumes various values from left to right along the £-axis, that is, for increasing values of x. The results may be tabulated conveniently as follows : X X 3 5a; 2 - 7x - 13 x s + 5x 2 — 7x — 13 - 10 - 1000 500 + 70 - 13 = 557 - 443 - 8 - 512 320 + 56 - 13 = 376 - 136 - 6 - 216 180 + 42 - 13 = 209 - 7 0+ - 13 = - 13 - 13 5 125 125 - 35 - 13 = 77 202 10 1000 500 - 70 - 13 = 417 1417 When x is negative, but numerically sufficiently large, it is seen that f(x) is negative. Thus, when x = - 10, x* = - 1000, while the sum of other terms, or the value of 5a; 2 - Ix - 13, is only 557 ; and /(-10) = -443. For points far enough to the right of the origin, say for x = 10, f(x) is positive. For example, /(10) = 1000 + 417 = 1417. Hence, the graph of y = f(x) would be located below the x-axis on the left but would rise above the #-axis as x gets larger and larger. As a second illustration, let us consider the function f(x) = x 4 + Ix 2 -8x + 10. Here f(x) is positive for large numerical values of x, regardless of the sign of x. Hence, in this case the graph Sec. 1 3-8 Theory of Equations 245 would be above the #-axis for large numerical values of x to both right and left of the origin. The following helpful conclusion can be drawn from the preced- ing discussion involving large values of x. When x is sufficiently large and positive, f(x) has the same sign as the leading coefficient a . When x is negative and sufficiently- large numerically, f(x) has the same sign as a when n is even, and has the opposite sign when n is odd. The following symbols are sometimes found in discussions of the values of functions for numerically large values of x. When the symbolic statement /(+oo)>0is used, what is meant is that, for all sufficiently large positive values of x, f(x) ig positive. Similarly, the statement /(+ oo ) < means that for all sufficiently large posi- tive values of x, f(x) is negative; the statement /(— oo) > means that, for all negative values of x which are sufficiently large numer- ically, f(x) is positive; and /(— co)< means that, for all negative values of x which are sufficiently large numerically, f(x) is negative. Thus, if f(x) =x* + 5x 2 -lx- 13,/(- oo) <0 while /(+ oo) > 0. However, if f{x) = x* + lx 2 - Sx + 10, /(- oo)>0and/(+ co)>0. 13-7. ROOTS BETWEEN a AND b IF f(o) AND f(b) HAVE OPPOSITE SIGNS Another helpful theorem relating to the roots of a polynomial equation is the following. Theorem. If the coefficients of a polynomial f(x) are real, and if a and b are real numbers such that f(a) and f(b) have opposite signs, then the equation f(x) =0 has at least one real root between a and b. We shall not give a proof of this statement here, but shall merely mention the following geometric considerations. The graph of a polynomial is a continuous curve; that is, it has no "breaks." There- fore, if the points (a,f(a)) and (&,/(&)) lie on opposite sides of the #-axis, the graph apparently has to cross the #-axis at least once between these points. 13-8. RATIONAL ROOTS The following theorem is fundamental for the solution of equa- tions having integral coefficients. Theorem, If the equation f(x) - a x n + a x x n ~ l H h a n - with integral coefficients has the rational root -, > where c and d a 246 Theory of Equations Sec. 13-8 are integers having no common factor > 1, then c is a divisor of the constant term a n , and d is a divisor of the leading coefficient a . Proof. We shall make use of a principle from the theory of num- bers: If an integer c divides the product of two integers a and b and if b and c have no common divisor other than ±1, then c is a divisor of a. Let -. be a root of f(x) = 0, where c and d are integers with no a common divisor other than ±1. Then pit s*n— 1 ** a °T« + ai d^ + '" + an ~ l d + a " = °- Multiplication by d n gives aoc n + a\c n ~ l d + • • • + a n ^icd n " 1 + a n d n = 0. Since c divides all terms before the final one, c also divides that term. If now c is factored into primes, none of these primes is a divisor of d, and therefore of d n . Thus each prime divides a n , and so c itself divides a n . In a similar fashion, it may be shown that d divides a . Example 13-7. Find the rational roots of Sx 3 - 17z 2 + 15z + 7 = 0. c Solution: The possible rational roots are of the form ~z > where c is a divisor of the constant 7 and d is a divisor of the coefficient 3. The only possible roots are ± 1, ± 7, ± 1/3, ± 7/3. Using synthetic division to check — 1, we obtain the following result: 3 - 17 15 7 1- 1 - 3 20 -35 3 -20 35 -28 By the remainder theorem, /( — 1) = — 28. Hence, — 1 is not a root. We see, however, that /(0) = 4-7. Therefore, by the theorem in Section 13-7, there must be a root between x = — 1 and x = 0. Examining our list of possible roots, in the hope that a rational root lies between — 1 and 0, we see that a pos- sibility is — 1/3. The check by synthetic division follows: 3 - 17 15 7 1- 1/3 -16-7 3 - 18 21 Hence, x = — 1/3 is a root. After dividing the given polynomial by x + 1/3, and equating the quotient Sx 2 - 18x -f 21 to 0, we obtain the quadratic x 2 - 6x + 7 = 0. This equation has the roots 3 ± s/2, which are not rational numbers. Therefore, the only rational root is x = - 1/3. In this example, as is sometimes the case, it has been possible to find all the roots of the given equation. Sec. 13-8 Theory of Equations 247 EXERCISE 13-4 1. Show that 18x 3 - 33a; 2 + 2x + 5 = has real roots between — 1 and 0, between and 1, and between 1 and 2. Find these three roots. 2. Find the rational roots of 2x 3 - 9x 2 + 3x + 4 = 0. 3. Prove that x 3 + 2x 2 - 3x - 5 = has at least one positive root. 4. Prove that x 4 - x 3 + x 2 + x - 3 = has at least one positive root and at least one negative root. 5. Prove the following corollary of the theorem in Section 13-8. If f(x) = x n + aix*- 1 + * • * + a n = has integral coefficients and has an integral root r, then r is a divisor of a„. 6. Find the integral roots of x 4 - 1 = 0. 7. Show that the equation x 2 + x + 1 = has no rational roots. 8. Solve the equation x 3 - 1 = 0. 9. Find all the integral roots of x 3 + x 2 + x + 1 = 0. 5 3 10. Show that 3 is a root of x 3 - -z x 2 - 2x + - = 0. Why does this not contra- dict the theorem in Section 13-8 or that in Problem 5? In each of the problems from 11 to 20, find all roots of the given equation. 11. x 4 - 8x 2 + 16 = 0. 12. 4x 3 - 16x 2 - 9x + 30 = 0. 13. 3x 3 + x 2 + x - 2 = 0. 14. 2x 3 + 3x 2 - 6x - 9 = 0. 15. 2x 3 - x 2 + 2x - 1 = 0. 16. x 3 - llx 2 + 37x - 35 = 0. 17. 5x 3 - 13x 2 + 16x - 6 = 0. 18. x 4 - 8x 3 + 37x 2 - 50x = 0. 19. 2x 3 - x 2 - 4x + 2 = 0. 20. 3x 3 - 13x 2 + 13x - 3 = 0. 14 Inequalities 14-1. INTRODUCTION In previous chapters we have explained some methods of deter- mining the roots of an equation. By applying these methods, one can find the values of an unknown for which a certain function of the unknown equals zero. Often, however, it is necessary to solve an inequality, that is, to discover for what values of the unknown a certain function is less than or greater than another function. The present chapter is concerned primarily with the solution of inequalities, and the following discussion is essentially an extension of the study of the order relation undertaken in Section 1-8. We, therefore, recommend that the student thoroughly review Section 1-8 before starting the study of the present chapter. Since the solu- tion of inequalities often involves the use of absolute values, a thor- ough mastery of Sections 1-9 and 1-10 is also a requirement. The classification of inequalities corresponds to that of equalities or equations. As in the study of equations, there are two kinds of inequalities involving unknowns, namely, absolute inequalities and conditional inequalities. An absolute inequality is an inequality that is satisfied by all values of the variable or variables for which the functions appear- ing are defined. A conditional inequality is one that is true only for certain values of the variable or variables. Thus, x 2 + 1 > (where x is real) is an absolute inequality, because it is true for every real value of x ; but x — 1 > is a conditional inequality, because it is valid only when x>l. 14-2. PROPERTIES OF INEQUALITIES The rules for dealing with inequalities are to some extent analo- gous to those for equations. In transforming inequalities, we shall have occasion to use the following elementary principles which 248 Sec. 14-3 Inequalities 249 follow at once from the fundamental properties proved in Section 1-8. Principle 1. If a < b, then a±c<6±c. Here are three illustrations : From 12 > 8, it follows that 12 + 3 > 8 + 3. From 8 < 12, it follows that 8 - 2 < 12 - 2. From 12 > 8, it follows that 12 - 8 > 0. n h Principle 2. If a < b and e> 0, then ac < be and - < - • c c Two illustrations are given here : From 3 < 5, it follows that 3 • 2 < 5 • 2. ft 10 From 8 < 10, it follows that £ < ~ • Principle 3. If a < b and c < 0, then ac > be and - > - • c c Here is an illustration : From 3 < 4, it follows that -3 > -4. 14-3. SOLUTION OF CONDITIONAL INEQUALITIES The process of solution of a conditional inequality consists in finding all values of the variable which satisfy the inequality. A solution consists of a set of values of the variable, rather than one or more isolated values as is usual in the case of a conditional equation. The discussion in this section is limited to inequalities involving rational functions in only one variable. If this variable is x, the inequality can be written in the form fix) > or f(x) < 0. For instance, suppose we want to solve the inequality x 2 - x > 2. We may then obtain the following equivalent inequality: /Or) = x 2 - x - 2 > 0. It is easily seen that this transformed inequality has the same solu- tion set as the original inequality. In solving this transformed inequality, we find first the values of x, if there are any, for which f(x) changes sign as x increases in magnitude. If f(x) is a polynomial, such a change of sign occurs when f(x) =0. In the example under consideration, changes of sign are obtained only at points where f{x) = x 2 - x - 2 = 0. Hence, we must find the roots of x 2 — x — 2 = 0. These are —1 and 2, and they determine on the #-axis three intervals throughout each of which f(x) retains the same sign. In other words, to find the 250 Inequalities Sec. 14-3 solution of f(x) > 0, we find the interval or intervals within which f(x) has the sign indicated in the given inequality. In the example under consideration this sign is positive, because f(x) is to be > 0. The method is applied in Example 14-1. In general, the solution of an inequality is obtained by equating the function f(x) to zero and solving the resulting equation. If the inequality is of the form ^-~ ^ 0, where p(x) and q(x) are poly- nomials, it may be cleared of fractions by multiplication by [q (x) ] 2 . Since the square of any non-zero real number is positive, the sense of the inequality is not changed by multiplication by this factor. This leads to the form f(x) ^ 0, where f(x) is a polynomial. The values of x for which f{x) changes sign are called critical values. When the critical values are arranged in increasing order, they determine on the #-axis intervals, throughout each of which f(x) cannot change sign. Consequently, the required solution is repre- sented by the set of values of x for which f(x) has the same sign as that indicated in the given inequality. Note. In general, it can be shown that if a factor of f(x) appears to an odd power (that is, if f(x) = has roots of odd multiplicity), the function will change sign at values of x for which this factor vanishes. If a factor of f(x) appears to an even power (that is, if f(x) = has roots of even multiplicity), the function will not change sign at values of x for which this factor vanishes. There- fore, it is sufficient to test only one value of x in one interval. This test gives the sign of f(x) in that interval. The sign of f(x) in each of the other intervals can be quickly and easily determined from the multiplicity of the critical values. Substituting into f(x) a value of x in each interval then provides a check of the solution. Example 14-1. Solve the inequality x 2 - x > 2. Solution: An equivalent inequality is f(x) = x 2 - x - 2 > 0. From the preceding discussion it follows that the critical values are the roots of x 2 - x - 2 = 0. These roots are — 1 and 2. As shown in Fig. 14-1, the points — 1 and 2 determine on the z-axis the following three intervals (a) x < - 1; (b) - 1 < x < 2; (c) x > 2. i ♦ i *■ ♦ 1 > X -2-10123 v „ / v M > v „ / (a) (b) (c) Fig. 14-1. Sec. 14-3 Inequalities 251 Throughout each of these intervals, f(x) retains the same sign. This condition may also be seen from the graph of y = x 2 — x — 2 shown in Fig. 14-2. Here we note that in each of these intervals the curve lies either entirely above the x-axis or entirely below it. The solution of f(x) > can now be found by examining the sign of /(#) in each of these intervals. Thus, for a value such as x = — 2 in the interval (a), /( — 2) = (- 2) 2 - (- 2) -2=4. Hence, f(x) is positive throughout the interval (a). In the graph of y = x 2 - x - 2 shown in Fig. 14-2, the curve lies above the x-axis to the left of x = — 1. For the value x = in the interval (6), we have /(0) = - 2. Hence, f(x) is nega- tive throughout this interval, and the graph lies below the x-axis. For the value x = 3 in the interval (c), /(3) = (3) 2 - (3) - 2 = 4. So/(x) is posi- tive, and the graph again lies above the .r-axis. These same results may also be obtained by the following much shorter procedure: Select a value of x in the interval (b) for which /(x) is easily evaluated. Such a value isx=O.Since/(0) = -2,/(x) <0 throughout this interval. Therefore, f(x) > for inter- vals (a) and (c), because the sign of f(x) changes at the critical values x = — 1 and x =2. We see, therefore, that in the intervals (a) and (c), f(x) has the sign indicated by the given inequality. Since f(x) must be greater than zero, the solution set of the given inequality is described by x < - 1 and x > 2. « *~X y-x 2 -x-2 Fig. 14-2. Example 14-2. Determine the values of x for which \/x 3 - 2x 2 - Sx is real. Solution: We shall solve the equivalent problem f(x) = x 3 - 2x 2 -3x^0. Solving the equation x 3 — 2x 2 — Sx = 0, we find that the critical values are — 1, 0, 3. fa -l o ■4- 3 -►X Fig. 14-3. (d) As shown in Fig. 14-3, these critical values determine on the x-axis the following four intervals: (a) x < ~1; (6) - 1 <s <0; (c) < x < 3; (d) x > 3. Throughout each interval /(x) has the same sign. In this example we may select x = 1 in the interval (c). Since /(l) =(1) 3 — 2(1) 2 - 3(1) = - 4, /(x) < throughout this interval. 252 Inequalities Sec. 14-3 As we proceed into the interval (6), we find that/(x) changes sign when x = 0. Therefore, f(x) > in the interval (6). Again /(x) changes sign when x = - 1 and becomes < in the interval (a). Proceeding to the right from the interval (c) into the interval (<2), we find that/Or) changes sign when x == 3 and is > in the interval (d). Thus, we have the following results: in (a), /(a) <0; in (&),/(«) > 0; in (c), f(x) < 0; in (d), /(x) > 0. Hence, the inequality /(x) > is satisfied in intervals (b) and (d). And, since the condition x 3 — 2x 2 — 3x = is also allowed in the original problem, the values — 1, 0, 3 are included in the solution. Therefore, the solutions for f(x) = x 3 — 2x 2 — 3x £ are — 1 £ x £ and x ^ 3. These are the values of x for which the original expression V# 3 — 2x 2 — 3x is real. x 3 — 3x 2 Example 14-3. What values of x satisfy ~- > 0? X — z Solution: Clear the given inequality of fractions by multiplying by (x — 2) 2 and obtain /(x) = {x - 2) (x 3 - 3x 2 ) > 0. Solving the equation x 2 (x - 2) (x - 3) = 0, we find that the critical values are 0, 0, 2, 3. Hence, we have the following four intervals, as shown in Fig. 14-4: (a) x < 0; (6) < x < 2; (c) 2 < x < 3; (d) x> 3. Throughout each of these intervals /(x) has the same sign. *»X Let us initially test/(x) for x = 1 in the interval (6). Since /(l) = (1 — 2) [(l) 3 - 3(1) 2 ] = ( - 1) ( - 2) = 2, it follows that/(x) > throughout this interval. We see that/(x) does not change sign for the critical value x = 0, because x = is a double root of /(x) = 0. Hence, /(x) > in the interval (a). As we proceed to the right from the interval (b) the function /(x) changes sign for each of the critical values x = 2 and x = 3. Hence, f(x) < in the interval (c), and/(x) > in the interval (fi). Therefore, /(x) is positive in the intervals (a), (6), and (d). That is, the solution set of the original inequality is described by x < 2, excluding the value x = 0, and x > 3. Example 14-4. Solve the inequality x 2 - 2x + 3 > 0. Solution: Let f(x) = x 2 — 2x + 3. Since the roots of f(x) = are imaginary, there are no critical values. Hence, the graph of y = /(x) lies either entirely above the x-axis or entirely below that axis. Testing for x = 0, we find that/(0) = 3. This result indicates that the graph lies above the x-axis. Consequently, the inequality is" satisfied for all real values of x. Sec. 14-3 Inequalities 253 I2x — 1 1 5 < 1. Solution: By Section 1-10, the inequality | x - 6 | < a is equivalent to I 2x — 1 I 6-a<x<6+a. Hence, — r — < 1, which is equivalent to | 2x — 1 1 < 3, I o I may be written as follows : 1 - 3 < 2x < 1 + 3 or - 2 < 2x < 4. We thus find that the solution set of the original inequality is described by - 1 < x < 2. Alternate Solution: We may proceed by solving individually the two inequalities 2x — 1 2x — \ - 1 < — r — and — - — < 1 and determining the common solutions; For 2x — 1 2x — 1 - 1 < — 5 — i we have - 3 < 2x - 1, or - 1 < x. For — r — < 1, we have 2x — 1 < 3, or x < 2. The common solutions satisfy the inequalities — 1 < x and x < 2. So we again have - 1 < £ < 2. EXERCISE 14-1 In each of the following problems, solve the given conditional inequality or inequalities. 1. x - 3 < 0. 2. x + 1 < 0. 3. x + 5 > 0. 4. x - 1 £ 0. 5. 4a; - 16 < 0. 6. 4x - 16 > 0. 7. 6x + 3 < 0. 8. 4x - 8 > 3x - 10. 9. 4s < - 3x - 7, 10. 3 - 4x < 2x -f 1. 11. - 3 < 6x < 3. 12. | x - 1 | < 3. M.0<* + 1 +1<1. u. _i<£^ + |<i. 15. | 2x - 3 | < 4. 16 -IHI< 2 - 17. z* < 169. 18. x 2 > 144. 19. 2x 2 ^ 32. »•!-!<«■ . 21. x(3x +2) < 1, 22. 4x 2 + 5x < - 1. «3<»-i- •24.H<-^-l. X 2 X 25. 3x 2 - 3x < - 4. 26. 3x 2 -f 6x < 9. 27. (x + 1) (x + 2) (x + 3) < 0. 28. (x - 4) (x + 5) (x - 6) > 0. 29. (x - 1) (x - 2) (x - 3) > 0. 30. (2x - 1) (x + 2) (3x + 1) ^ 0. 31. y/x 2 - 25 is real. 32. y/x 2 - 5x + 6 is real. 33. *&£!>! > 0. 34. 4/^T < 1- x + 2 r x + 1 35. |2L±!| < i. 36. -; £ r <0. Ixl x 2 — 2x - 3 . 254 Inequalities Sec. 14-4 14-4. ABSOLUTE INEQUALITIES To prove the truth of an absolute inequality, one must use the known properties of the order relation. When none of these seems readily applicable to the given inequality, it may be helpful to replace this inequality by an equivalent one which may be more easily treated. Repeated replacements may have to be made. In carrying out a sequence of replacements, one need not verify equivalence at each stage, provided that the final inequality can be shown to imply the original one. The methods for proving absolute inequalities may be used also to prove theorems involving inequalities, as in Example 14-7. Example 14-6. Prove that a 2 + b 2 ^ 2ab for all real numbers a and b. Solution: The given inequality is equivalent, by Principle 1, Section 14-2, to a 2 - 2ab + b 2 £ 0, that is, to (a - b) 2 £ 0. This last inequality is true, because the square of every real number is non- negative. Therefore, the original inequality is true also. a c Example 14-7. If a, b, c, and d are distinct positive real numbers, and if -r < -? > prove that l<rT1 < r Solution: The inequality t < , , is equivalent, by Principle 2, Section 14-2, to ab -f- ad < ab + be. This inequality is equivalent, by Principle 1, to ad < be. a c This inequality is true because it is equivalent to the given condition r < -i by Principle 2. Therefore, the inequality t < , , is true also. Similarly, the inequality , , < -j is equivalent, by Principle 2, to ad + ed < be + ed, and this inequality, by Principle 1, is equivalent to ad < be. a c This last inequality again is equivalent to the given condition r <~j* Hence, a A~ c c the inequality , , < -j is true also. Therefore, it follows that the original ,.,. a a + c c . inequalities t < , , < -j are true. Sec. 14-4 Inequalities 255 EXERCISE 14-2 1. Prove that a 2 + 1 ^ 2a if a is real. Note that a 2 + 1 = 2a only when a = 1. 2. Prove that ^4^ ^ "^TT if a > and 6 > 0. 2 a + 6 3. Prove that a 2 > a if a > 1 ; and that a 2 < a if < a < 1. 4. Prove that a > a 3 if < a < 1. 5. Prove that t + - > 2 if a and 6 are positive and a j* b. o a 6. Prove that a 2 +b 2 + c 2 ^ ab +bc + ca. (Hint: From Example 14-6, a 2 + b 2 £ 2a&, 6 2 + c 2 ^ 26c, and c 2 + a 2 £ 2ca. Add these inequalities.) 7. If a and b are two positive real numbers, the quantities — - — > \fab, and —tj- are called, respectively, the arithmetic mean (A), the geometric mean (G), and the harmonic mean (H) of a and b. Prove that H < G < A, except when a = 6. (In this case, we have A = G = //.) 8. Prove that a 3 + & 3 > 3afr(a — 6), if a and 6 are positive and a > b. 9. Prove that ab + cd g 1, if a, 6, c, and d are positive, and if a 2 + & 2 = 1 and c 2 + d 2 = 1. 10. Prove that 7— ; — ^ t » according as a $ b> if a, 6, and c are positive. o + c b 11. Prove that a*b + ab 3 < a 4 + 6 4 , if a ^ 6. 12. If z = a; + yi is a c omplex n umber, the modulus or absolute value of z is denoted by | z \ = y/x 2 -f- t/ 2 . Show that | z x + z 2 \ £\zi\ +\z 2 \ and that I zi — z 2 1 ^ | 2i | — | 32 | , for all complex numbers Z\ and z 2 . 13. If Zi, z 2 , and Z3 are complex numbers, prove that | Z\ +z 2 1 ^ | Z\ +Z* | -f |z 2 —z% |. 14. If z is a complex number, prove that ^ | | 2 | + z\ < 2 | z |. 15 Progressions 15-1. SEQUENCES AND SERIES Sequences. An infinite sequence, called more simply a sequence or sometimes a progression, is a single-valued function whose domain of definition is the set of positive integers. A finite sequence is a single-valued function whose domain consists of the integers 1, 2, - - - , m for some positive integer m. In specifying a sequence, it is necessary to give a definite rule of correspondence which assigns to each integer n a single definite number, or term, of the sequence. This term may be denoted by a n . In particular, there is a first term ai corresponding to the integer 1, a second term a 2 corresponding to the integer 2, and so on. The sequence may be specified by the array of numbers (15-1) ai, d2, az, • • • , a n , • • • . This sequence is denoted briefly by \a n ). (As usual, throughout this discussion, a row of dots • • • stands for numbers assumed to be present but not written.) A finite sequence has a "last" term and may be designated by a l9 a- 2 • • • , a m or simply by {a w }. The nth term, or general term, of a sequence is denoted by a n . From the rule specifying the nth term for each n, we obtain the first, second, third, and other terms of the sequence by substituting for n the values 1, 2, 3, and so on in turn. For example, if a n = 1/n, the sequence is 111 We should note that {a n } is the symbol for the sequence or func- tion as a whole, whereas a n is the symbol for the nth term or value of the function corresponding to the integer n. 256 Sec. 15-1 Progressions 257 There are many methods for specifying the function in the defini- tion of a sequence. Two of these methods follow : Explicit Formula, In one method, the nth term is given in terms of n itself by means of an explicit formula. Here are a few illustrations. If a n = ft, the sequence is 1, 2, 3, • • • . T r n ,, .12 3 If a n = 9 , , > the sequence is - > ■= > 77: > • • • . rr + 1 2 5 10 xr n xi. .,23 If a n = ~ 7 > the sequence is 1, - > « > • • • . Zti — 1 * o o Recursion Rule. In another method, one or more of the first sev- eral values of a n are given explicitly, and a rule is then given whereby a n can be calculated from some or all of its predecessors. A few illustrations are given here. Let a n+ i = a n 2 + 1 with a\ = 0. Then a 2 = ai 2 + 1 = 2 + 1 = 1, a s = a 2 2 + 1 = l 2 + 1 = 2, a 4 = a 3 2 + 1 = 2 2 + 1 = 5, Let a n+ i = (n + l)a n with 01 = 1. Then a 2 = 2ai = 2 • 1 = 2, a 3 = 3a 2 = 3 • 2 = 6, a 4 = 4a 3 = 4 • 6 = 24, Note that in this example a n = n! Let a n +2 = «n+i + «n with a\ = and a 2 = 1. Then a 3 = a 2 + ai = 1 + = 1, a 4 = a 3 + a 2 = 1 + 1 = 2, a 5 = a 4 + a 3 = 2 + 1 = 3, Series. Let {a n } be a given sequence of terms ai, 03, • • • , a*, • • ■ . Form a new sequence {s n }, where s n is obtained by adding the first n terms of \a n }. The sequence of partial sums is then given by $1 = aij S2 = ai + a 2 , • • • , Sn = ai + a 2 H (- a n , • • • (n = 1, 2, 3, • • •)• The sequence {s n } formed in this way is called the (infinite) series based on the given sequence {a n }. The series as just defined is usually written in the following abbreviated form : (15-2) ai + a 2 + • • • +On + • • • . 258 Progressions Sec. 15-1 Two illustrations of series follow. Illustration 1. Let a n = - and s n = ai + a% H I- 0n. Then the n partial sums are given by Sl = ai = 1, «2 = &1 + «2 = 1 + o = o ' J_ _1_ i.l.l 11 S3=al + a2 + a3 = l+2 + 3 ::=: "5' , The series may then be written 1 + i + i H + =• + n 2 + w ' Illustration 2. Let a n = -o-; — and s n = «i + &2 + • • • + a n . Then the partial sums are given by The series is s ± = a x = - , j_ 1.1 2 52 = ai + a2 = o + g = « > _i_ . 1,1,1 3 «3 =ai + a2 + «3 = 2612 = 4 : I + I + i, . .. + _!_ + 2 ^ 6 ^ 12 ^ n 2 + r* Limit of a Sequence. One of the most important questions relat- ing to sequences is whether or not a given sequence { a n } has a limit as n increases indefinitely. If such a limit <£ exists, the sequence is said to be convergent or to converge to the limit <£. Symbolically this statement may be expressed as follows : lim a n = £. n-»oo This notation is read, "the limit of a n as n increases indefinitely is £." In order to help make the concept of limit clear, we shall consider the sequence given by a n = - • The terms are n 1 1 1 a\ = 1, a2 = g > as = ^ > • • • > a n = - > • • • . When we examine these terms, we notice that the larger the number n is, the smaller the term becomes. In other words, as n increases, the closer to zero is the number - • In fact, - can be made as small n n as we please, if we merely choose n sufficiently large. For example, Sec. 15-1 Progressions 259 - < — ; - for every n larger than 100. We see also that - < n 100 * n 1000 for every n larger than 1000. We may conclude that for an arbi- trary real positive number d, we can find a value of n, say any integer N ^ -j > such that for all integers n> N, it is true that - < d. The limit and the sequence I - \ are therefore related as indicated by the following statement. The sequence j - > converges to the limit <£ = 0, if to each arbi- trary positive number d there corresponds an integer N > such that — d < - < + d f or every n > N. n In general, the limit £ of a sequence may be defined as follows : Definition of Limit of Sequence. A sequence {a n } converges to the limit «£, if to each arbitrary real number d > 0, there corresponds a positive integer N such that £— d < a n < £ + d for every n> N. This definition may also be put as follows : Definition. A sequence { a n ) converges to the limit <£, if for each number d > there exists a positive integer N such that | £ — a n | < d for every n > N. Convergence of a Series. We shall again consider the infinite series (15-2) ai + a 2 H h a n -) , which is the sequence of partial sums «li «2i • • • , «m • • • of the sequence {a n }. By the following definition the series is con- vergent if the sequence of partial sums is convergent. Definition. If the sequence of partial sums of the infinite series (15-2) converges to a limit, and if lim S n = S, then the series is said to converge to the limit S, and S is called the sum of the infinite series. The new use of the word "sum" for the value S of an infinite series is perhaps unfortunate, for it seems meaningless to talk about adding up the terms of an infinite series. Actually, S is not a sum, but it is rather the limit of a sequence of partial sums of the series. If a series does not converge to a limit as n becomes infinite, we say that it is divergent, or that it diverges. 260 Progressions Sec. 15-1 EXERCISE 15-1 1. Given a x = 1 and a n +i = ft + a„. Find the five terms of the sequence {a n }. 2. Given a x = 4, a 2 = 3, and a„ +2 = 2a„ +l — a„. Find the first six terms of the sequence { a n } . 3. Given a x = 1, a 2 = 2, and a„ +£ = (n + l)a n+ i — na n . Find the first six terms of the sequence { a n } . 4. Show that the sequence a n =■ 3 n satisfies the recursion rule a n+ 2 = cin+i + 6a n . 5. Show that the sequence a n = (— 2) w also satisfies the recursion relationship of Problem 4. 6. Assuming that A and B are any real numbers whatsoever, show that the sequence a n = A3 n + B{ - 2) n satisfies the relationship of Problem 4. 7. Given a n+ i = 2a n + 1 with a x = 2. Let s n = «i + a2 + • • • + a„. Find fli, Q>2, ' - • , ct5 and Si, s 2 , • • • , «s. 8. Show that if s ft = ai + «2 + • • • + «», then s n+ i — s n = a n +i. 9. Prove that a n = s» — s»_i for n > 1, given ai = s 1# 10. If s n = ft 2 , show that a n = 2ft — 1 and therefore that 1 + 3 + 5 + • • • + (2ft - 1) = ft 2 . 15-2. ARITHMETIC PROGRESSIONS An arithmetic progression is a sequence of numbers in which each term after the first is obtained from the preceding one by adding to it a fixed number; this number is called the common difference. Note that the common difference may be found by subtracting any term of the sequence from the one that follows. Thus, 1, 5/2, 4, 11/2, is an arithmetic progression with the common difference 3/2, since 11/2 -4 = 4-5/2 = 5/2-1 = 3/2. Also, 5, 1, -3, -7 is an arithmetic progression with the common difference —4, since -7- (-3) =--3-1 = i_5 = -4. It follows, therefore, that a necessary and sufficient condition that three numbers A, B, and C form an arithmetic progression is C-B = B-A. 15-3. THE GENERAL TERM OF AN ARITHMETIC PROGRESSION Let a denote the first term of an arithmetic progression, and let d denote the common difference. Then, by definition, an arithmetic progression with n terms may be written as follows : a, a + dj a + 2d, a + 3d, • • • , a + (n — l)d. Hence if l n represents the value of the nth term, (15-3) l n = a + (n - l)d. Sec. 15-4 Progressions 261 Also, we may write an arithmetic progression of n terms in the following manner : a, a + d, a + 2d, • • • , l n — 2d, Z n — d, l n . We shall be concerned with five quantities in connection with an arithmetic progression. These are the first term a, the number of terms n, the nth term l n , the difference d, and the sum S n of the n terms. 15-4. SUM OF THE FIRST n TERMS OF AN ARITHMETIC PROGRESSION Let S n represent the sum of the first n terms of an arithmetic progression. If we write the indicated sum in both direct and reverse orders, we have S n = a + (a + d) + (a + 2d) + • • • + (l n - 2d) + (l n - d) + l n , and S n = l n + (k - d) + (I» - 2d) + • • • + (a + 2d) + (a + d) + a. Adding the right sides, we have n terms each of which is a + l n . Thus, 2S» = (a + l n ) + (a + W) + • • • + (a + t) + (a + l n ) = n(a + J n ). Therefore, (15-4) S» = | (a + In). If we substitute a + (n - l)d from (15-3) for l n in (15-4), we have another useful form for the sum. This is (15-5) S» = | [2a + (n - l)dj. Example 15-1. Determine which of the following sequences are arithmetic progressions: a) 3, 7, 10; b) 6, 1, - 4; c) 3x - y, 4z + 2/, 5z + 3y. Solution: a) Since the differences 10-7 and 7 - 3 are not equal, the sequence 3, 7, 10 is not an arithmetic progression. b) In the second sequence, -4-1=1-6=- 5. Since these differences are equal, the sequence 6, 1, — 4 is an arithmetic progression. c) We find that (5x + Zy) - (4x + y) = x + 2y and (4z + #) - (3.r - y) = £ +2?/. Since there is a common difference, the given sequence is an arithmetic progression. Example 15-2. Find the twelfth term, and also the sum of the first 12 terms, of the arithmetic progression 4, 7, 10, • • ■ . 262 Progressions Sec. 15-4 Solution: We have a = 4, n = 12, and d = 3. Then, by (15-3), l 12 = a + (n - l)d = 4 + (12 - 1)3 = 37. Also, by (15-4), flu = \ (a + I.) = y (4 + 37) = 246. Example 15-3. The third term of an arithmetic progression is 3/4, and the sixth term is 3/2. Find the twenty-second term. Solution: By (15-3), h = a -f 2d and U = a + 5d. Thus, we have f a + 2d = 3/4, \a + 5d = 3/2. By solving these two linear equations, we find that a = 1/4 and d = 1/4. By (15-3), I22 = 1/4 + 21(1/4) = 11/2. Example 15-4. Find each value of x for which the three quantities Zx — 5, x + 4, Sx — 2 form an arithmetic progression. Solution: Applying the condition C - B = B — A, we have (3a: - 2) - (x + 4) = (x + 4) - (3x - 5). o 1 • u- 15 ^ ... .. . . 25 31 37 Solving, we obtain a* = -j- • The arithmetic progression is -j- > -j- » -j- • 15-5. ARITHMETIC MEANS The terms of an arithmetic progression between any two given terms are called arithmetic means between the given terms. If we let the given terms be a and l H , any number, say k, of means may be inserted between a and l n by using the formula l n = a + (n — l)d with n — k + 2. As soon as we have found d, we can insert the required means. Example 15-5. Insert three arithmetic means between and 1 . Solution: Let a = 0, l n = 1, and n = 3 + 2 = 5. Then 1 = + 4d, and d = | • Hence, the three means are 7 » x » 7 • 4 2 4 If only a single arithmetic mean is to be inserted between two given numbers, then the inserted value is called the arithmetic mean of the given numbers. Thus, if a, x, b form an arithmetic progression, x is called the arithmetic mean of a and 6. Since b — x = x — a, a + 6 Thus, the single arithmetic mean of two numbers is equal to one- half their sum. Sec. 1 5-5 Progressions 263 EXERCISE 15-2 In each of the problems from 1 to 9, determine if the given sequence is an arith- metic progression. Find the next two terms of the extension of each arithmetic progression. 1. 5, 8, 11, 14. 2. - 1, 7, 13, 19. 3. - 10, - 3, 4, 11. 4.4,12,19,27. «.-H'l.J- ••t'B'-i'-IB- 7. a, 6, 26 - a, 36 - 2a. 8. a + 6, a — 6, a — 26, a - 36. 9. — - — > a, — ^— • In each of the problems from 10 to 19, find l n and S n for the arithmetic progression. 10. 2, 8, 14, • • • to 12 terms. 11. 3, 6, 9, • • • to 26 terms. 12. 22, 18, 14, • • • to 7 terms. 13. 1, 2, 3, • • • to 10 terms. 14. 2, 4, 6, • • • to 50 terms. 15. 1, 3, 5, • • • to 75 terms. 16. a, 2a, 3a, • • • to 10 terms. 17. 0.2, 0.5, 0.8, • • • to 20 terms. 18. 1, 8, 15, • • • to 35 terms. 19. 1, 2, 3, • • • to 100 terms. In each of the problems from 20 to 27, three quantities relating to an arithmetic progression are given. Find the other two quantities. 20. a = 5, h = 36, n = 4. 21. a = 10, n = 10, d = 10. 22. S 2 x = 653, n = 21, a = 6. 23. n = 45, d = | , S45 = 63. 24. Z21 = 8, n = 21, d = ^ • 25. Z» = 7, & = 52, d = | • 1 3 IS?*! 26. a = - i , d = ~ , S n = ^- 27. a = 1, S. = 45, d = 1. 28. Find the sum of the first 100 even integers. 29. Find the sum of the first n odd integers. 30. Insert five arithmetic means between 20 and 30. 31. Insert ten arithmetic means between 100 and 40. 32. Insert six arithmetic means between — 3 and — 2. 33. Find the arithmetic mean of 10 and 56 and that of 4 and 28. 34. Find the arithmetic mean of 28 and 65 and that of 33 and 78. 35. Insert k arithmetic means between - and a, where a^0, a 36. A display of cans in a grocery store is in the form of a pyramid whose base is an equilateral triangle. If each side of the base contains 20 cans and the number of cans decreases by one for each successive row, how many cans are in the display? 37. A lottery contains tickets numbered consecutively from 1 to 100. Customers draw tickets and pay according to the number of the ticket, except that tickets numbered above 50 cost just 50 cents each. How much money isj collected if all tickets are sold? 38. Determine z so that x, x - 2, Sx will be an arithmetic progression. 39. The sum of the first and fourth terms of an arithmetic progression is 20. The sum of the third and twelfth terms is 36. Find the sum of the first 15 terms. 40. Find the sum of all multiples of 5 from 100 to 1,000, inclusive. 264 Progressions Sec. 1 5-6 15-6. HARMONIC PROGRESSIONS A harmonic progression is a sequence of non-zero numbers whose reciprocals form an arithmetic progression. Thus, a, b, c are in harmonic progression if - > r > - form an arithmetic progression. a b c The terms of a harmonic progression between any two given terms are harmonic means between the given terms. To insert a desired number of harmonic means between two numbers, we insert the same number of arithmetic means between the reciprocals of the two given numbers and then invert the result- ing terms. The harmonic mean of two numbers is found in the follow- ing manner. If a, x, b form a harmonic progression, then - > - > t a Qu o form an arithmetic progression, and - is the arithmetic mean of - I x a and r • Hence, b 1 « b Solution of this equation for x yields the harmonic mean 2ab a + b Clearly the harmonic mean exists only if a + b ¥= 0. Example 15-6. Insert three harmonic means between — 3 and 2. Solution: The corresponding arithmetic progression is — ■= t • • • > -z • Here a= — ^ > 1 11 5 h = - i and n = 5. Hence, - = — - -f 4d, and d = ~j • It follows that the arithmetic progression is — - > — o ' To > o7 ' o ' 24 Therefore, the three harmonic means are — 8, 12, -=- • EXERCISE 15-3 In each of the problems from 1 to 6, determine if the given sequence is a harmonic progression. Find the next two terms of the extension of each harmonic progression. t 1 1 JL 9 111 .111 *" 3'7' if A 4'8'16' 6u 5 , T0 , 15 # *• I'!'!- 5. 16,8,^. «.-2,2,f. 7. Find the tenth term of the harmonic progression - > = > — ' * " ' • 8. Find the seventh term of the harmonic progression 6, 3, 2, • • • . 9. Insert four harmonic means between 1 and 2. Sec. 15-8 Progressions 265 10. Insert three harmonic means between ~ and j • 11. Insert four harmonic means between 6 and 24. 12. Find the harmonic mean of 6 and 9. 13. Find the harmonic mean of 24 and 72. 2 3 14. Insert nine harmonic means between - and ~ • o Z 15. If a 2 , b 2 , c 2 form an arithmetic progression, show that b -f- c, a + c, a + b form a harmonic progression. 16. If a, b, c form an arithmetic progression and b, c, d form a harmonic progression, show that ad = be. 17. If x is the harmonic mean of a and b, show that 1 r = — h t • x — a x — b a b 18. If a, 6, c, d form a harmonic progression, show that —i = __ , • 19. If a, 6, c form a harmonic progression, show that - = r • 20. If the harmonic mean of a and 6 is equal to their arithmetic mean, show that a = b y and conversely. 15-7. GEOMETRIC PROGRESSION A geometric progression is a sequence of numbers in which each term after the first is obtained from the preceding one by multi- plying it by a fixed number; the multiplier is called the common ratio. The common ratio may be found by dividing any term by the one immediately preceding it. Thus, 1, - > - > ^ is a geometric progression in which the common ratio is ?:-:-t : = : t- ; -^ = ^- j -1=o' Also, 8 4 4 2 2 2 \/2, 1, —.= i - is a geometric progression in which the common ratio y2 2 is - -*- — = = -7= -*- 1 = 1 -5- V 2 = — =• 2 V2 \/2 V2 It follows that a necessary and sufficient condition that three non- C B zero numbers A, B, and C form a geometric progression is -» = -j • 15-8. THE GENERAL TERM OF A GEOMETRIC PROGRESSION Let a denote the first term of a geometric progression, and let r denote the common ratio. Then the progression may be written as follows : a, ar, ar 2 } ar 3 , • • • , ar n ~ l . Hence, if l n represents the value of the nth term, (15-6) l n = ar*~ l 266 Progressions Sec. 15-9 15-9. SUM OF THE FIRST n TERMS OF A GEOMETRIC PROGRESSION Let S n represent the sum of the first n terms of a geometric progression. Then S n = a + ar + ar 2 + • • • + ar n ~ 2 + ar n ~~ l . Multiplying by r, we have S n r = ar + ar 2 + ar 3 + • • • + ar 11 " 1 + ar n . Subtracting the first of these equations from the second, term by term, we have S n r — S n = ar n — a. Therefore, (15-7) 5n = o(^_l) = a(^) ( ^ 1} If we multiply both sides of (15-6) by r, we get rl n = ar n . Sub- stituting in (15-7), we obtain another useful form for S n . This is (15-8) S n = ^^ (r ^ 1). Note. If r = 1, these formulas do not apply; in this case, however, the geometric progression becomes a + aH haton terms, and S n = na. Example 15-7. Determine which of the following sequences are geometric progressions: a) 2,6,18; b) 5,10,30; c) x,j>~- Solution: a) The ratios found by dividing each of the second and third terms by the preceding one are 3 and 3. Hence, the sequence 2, 6, 18 is a geometric progression. 6) In this sequence, the ratios of consecutive terms are 2 and 3. Therefore, the sequence 5, 10, 30 is not a geometric progression. c) The given sequence is a geometric progression in which the common ratio is x/y. Example 15-8. Find U and Ss in the geometric progression 6, 2/3, 2/27, Solution: The common ratio is (2/3) -s- 6 = 1/9. Since a = 6 and n = 5, we have l 5 = ar*- 1 = 6(l/9) 4 = ■ 6 ai 65 6! ( 1 \ „ _ a(l - r») _ 6(1 - (1/9)*) _ 6 \ 1 59,049/ _ 14,762 6 1 - r 1-1/9 ~~ 1-1/9 2187 Example 15-9. The fifth term of a geometric progression is 3, and the tenth term is — 96. Find the common ratio and the first term. Solution: By the formula (15-6) for the nth term of a geometric progression, we have ar* = 3 and ar g = - 96. Sec. 15-10 Progressions 267 Dividing each side of the second equation by the corresponding member of the first equation, we obtain r 6 = — 32. Hence, r = - 2. Therefore, a( - 2) 4 = 3, or 3 Example 15-10. Find each value of x for which the three numbers x, x — 2, x + 1 form a geometric progression. C B x 4- \ x — 2 Solution: If we apply the condition -5- = -j 1 we have ^ = • Solving, Jj A X — m Z X we obtain x =4/5. Hence, the geometric progression is 4/5, — 6/5, 9/5. 15-10. GEOMETRIC MEANS Terms of a geometric progression between any two given num- bers are called geometric means between the given numbers. Let a and l n be given numbers. Then k means may be inserted between them by using the formula l n = ar*- 1 with n = k + 2. Example 15-11. Insert three geometric means between 1 and 2. Solution: Let a = 1, l n = 2, and n = 5. Then Z 5 = cm* 4 and r = 2 1/4 . Hence, the three means are 2 1 ' 4 , 2 1 ' 2 , 2 3 ' 4 . If a, x, b form a geometric progression, then x is called a greo- b x metric mean of a and 6. Since - = - » a; a x = db V^- Hence, we note that a geometric mean of two numbers is the same as a mean proportional between the two numbers. EXERCISE 15-4 In each of the problems from 1 to 9, determine whether the given sequence is a geometric progression. Find the next two terms of the extension of each geometric progression. 1. 2, 8, 32. 2. ! , - 1, 2. 3. 4, 16, 64. 4. 2,4,6. 5. 27,18,12. G# ^'i' A* _ V3 V6 V3 k n a* a\ Q 1 1 1 In each of the problems from 10 to 15, find l n and £ n in the given geometric progression. 10. 4, 2, 1, • • • ; n = 10.. 11. 3, 2, 4/3, • • • ; n = 15. 12. 3, 9, 27, • • • ; n = 45. 13. 100, - 10, 1, • • • ; n = 101. 14. log 2, log 4, log 16, • • • ; n = 10. 15. log 9, log 3, log \/3, • • • ; n = 6. 16. Find the sum of the first n terms of the geometric progression 1, ^ > j > • • • . 268 Progressions Sec. 15-10 17. For what values of x do x - 2, x — 6, 2x + 3 form a geometric progression? 18. For what values of x do 3x -f 4, a; — 2, 5z -f 1 form a geometric progression? 19. For what values of x is x -f 1 a geometric mean of 2x + 1 and x — 1? 20. Find a geometric mean of 4 and 16. Also, find their arithmetic mean. 21. Find a geometric mean and the arithmetic mean of 3 and 12. 22. Insert four geometric means between 1 and 32. 23. Insert five geometric means between 1 and 1,000,000. 24. Insert ten geometric means between 1 and 2. 25. If A, G, and H denote the arithmetic mean, a geometric mean, and the harmonic mean, respectively, of two numbers a and 6, prove that G 2 = AH. 26. If the arithmetic mean of a and 6, when a, b ^ 0, is A, a geometric mean is G, and the harmonic mean is 77, find A — G,G — H, and A — H. 2 4 27. Show that the sum of the first n terms of the geometric progression 1, ^ > ^ > • • • is 3(1 — (o))* Discuss how this sum varies as n increases. 28. Show that the sum of the first n terms of the geometric progression 8, 4, 2, • • • is 16 (l — (o) ) ' Discuss how this sum varies as n increases. 29. Find the sum of the first n terms of the sequence 1, 2x } 3x 2 , 4a: 3 , • • • . (Hint: Let S n be the sum. Then compute S n — xS n .) 30. Find the sum of the terms of the finite sequence 2, = > =^ > -=^ > • • • > — =^ — • 15-11. INFINITE GEOMETRIC PROGRESSION A geometric progression in which the number of terms is infinite is called an infinite geometric progression. In Section 15-9, we found an expression for the nth partial sum S n of a geometric progression. Hence, S n is the nth term of the geometric series based on the given progression. Thus, we have Si = a; S2 = a + ar; S3 = a + ar + ar 2 ; • • • ; S n = a + ar + • • • + ar"- 1 . Furthermore, „ _ q(l - r») , . s n - 1 _ r fr * 1), or 1 — r 1 — r Let us consider what happens to the sum of n terms of a geo- metric progression when the number of terms increases indefinitely. If a = 0, evidently S n = whether r ¥" 1 or r = 1. In this case, the number meets the requirement of a limit of S n , so that the sum S of the infinite geometric series is equal to 0. Now suppose a ¥* 0. For this condition, we consider four cases. Case 1. Assume that \r\ < 1. If r ^ 0, the numerical value of r* decreases as n increases. Moreover, by making the number of terms Sec. 15-12 Progressions 269 sufficiently large, we can make \r n \ as small as we please. It follows that if \r\ < 1, we can make S n differ from by as little as we 1 — T please ; that is, S n approaches -- as a limit. This condition may be stated symbolically in the following manner: (15-9) 8 = lim S n = a 1 - r where S is the sum of the infinite geometric series. Equation (15-9) is true also if r = 0, since in this case S n has the constant value a. Case 2. If r = 1, then S n = na. Since a ¥" 0, |S„| increases indefi- nitely as n increases. Here { S n J diverges. Case 3. If \r\ > 1, then Ir 71 ] increases indefinitely as n increases. fl (lT n Hence, so does I S n I = I z z I • Again {S n } diverges. 1 — r 1 — r Case I>. If r = — 1, then the progression becomes, a, —a, a, —a, • • • , (— l) w x a. If n is even, S n = 0. If n is odd, S n = a. In this case, we say that S n oscillates between and a. Here also the series diverges. The sum of an infinite geometric progression can therefore be found by (15-9), but only when \r\ < 1. When r = 1, r — — 1, or |r| > 1, the series diverges, and we say that the series has no sum. For a further discussion of this topic, the student may refer to a treatise on the theory of limits. Example 15-12. The owner of a fleet of trucks finds that if used motor oil is refined for re-use, 20 per cent of the oil is lost in the process. If he starts with 100 gallons of refined oil and re-refines this oil each time it becomes dirty, determine the total amount of oil he has used before the entire 100 gallons is lost. Solution: We begin with 100 gallons. After the first reclaiming operation, we have 80 gallons of good oil. When this becomes used and is re-refined, we have 64 gallons; and so on. Theoretically, we would never use up the entire amount of oil. However, the limit of the sum of the amounts of oil reclaimed is approximately reached after a large number of operations. Hence, we have an infinite geometric progression in which a = 100 and r = 4/5. Then S = jyz = 500. Thus, by re-refining the oil as it is used, the fleet owner has had the equivalent of 500 gallons of oil. 15-12. REPEATING DECIMALS If a decimal contains a fixed sequence of digits which are repeated indefinitely, we call it a repeating decimal. Thus, 0.135135 • • is a repeating decimal. This decimal is written 0.i35, the dots indicating the first and last digits of the sequence which is to be repeated. Also, 0.34516 = 0.34516516516 • • • . A repeating decimal is wholly or partly an infinite geometric series. For 270 Progressions Sec. 15-12 example, since 0.34516 = 0.34 + .00516 + 0.00000516 + ••• , it is composed of the decimal 0.34 and an infinite geometric series in which a = 0.00516 and r = 0.001. Hence, since \r\ < 1, n qakia -hqa± 000516 _ ft Q , , 0.00516 _ 34 172 _ 11494 Ud451b " Ud4+ l -0.001 "" Ud4+ 0.999 " 100 + 33300 "~ 33300 ' Note that if we divide 172 by 33300, we obtain the repeating deci- mal 0.00516. Example 15-13. Express the repeating decimal 0.263 as an equivalent numerical fraction. Solution: We can write this decimal in the form 0.2 + 0.063 + 0.00063 + • • • . Hence, the required number consists of the decimal 0.2 plus an infinite geometric series in which a = 0.063 and r = 0.01. The sum of the series, since \r\ < 1, is 0.063 0.063 7 8 = 1 - 0.01 0.99 110 1 7 9Q Therefore, 0.263 = ± + jjg = f± • EXERCISE 15-5 In each of the problems from 1 to 12, find the sum of the convergent series based on the given infinite geometric progression. 1. ^ > — 1, g ) • • • . 2. g > - > 2= > • • • . 3. 3, \/3, 1, • • • . 4. — * -r > ~ i • • • • 5. 8, 4, 2, • • • • 6. 8, — 4, 2, • • • • 24 a 112 q i 3 9 10. 0.5, 0.05, 0.005, • • • . 11. 0.18, 0.0018, 0.000018, 12. 0.3 + 0.012 + 0.00012 + 0.0000012 + • ■ • . 13. Find the sum of the series based on the infinite geometric progression 24, 8 8, 5 ) - • • . Also, find the sum of the first 20 terms of this progression and com- o pute the error introduced by using S instead of $ 2 o. 14. What would be the error if S were used instead of S n for the sum of the geometric progression 48, — 36, 27, • • • to 10 terms? 15. A ball is dropped from a height of 3 feet. On each rebound it bounces back to three-fourths the height from which it last fell. Assuming that this bouncing continues indefinitely, find the distance it travels in coming to rest. How far has it traveled after bouncing ten times? 16. A swinging pendulum will gradually come to rest as a result of friction. If, on each upswing, the pendulum swings through 98 per cent of the arc through which it fell, and if the initial arc for one complete swing was 20 inches, find the distance traveled before the pendulum comes to rest. Sec. 15-13 Progressions 271 Convert each of the following repeating decimals to fractional form. 17. 0.1. 18. 0.15. 19. 0.90. 20. 0.243. 21. 0.16. 22. 0J42857. 23. 2.9. 24. U234, 25. 0.11542. 26. 2.i23. 15-13. THE BINOMIAL SERIES We shall now consider the binomial expansion when n is any real number. If we let a = 1 and 6 = x in (4-13), the binomial formula becomes (15-10) (i+ g ). = i + M + 5^* + n(n - 1) ( n - 2) ■ ■ ■ (n - r + 1) + • • • H -; x r + • • • . r! The right member of (15-10) is called a binomial series. We saw in Section 4-6 that, if n is a positive integer, the series on the right in (15-10) terminates with x n , and (15-10) is true. Otherwise, the terms of the series continue indefinitely, giving rise to an infinite series. The question which now arises is whether (15-10) is valid when n is not a positive integer ; that is, whether the series on the right converges, and, if so, whether its sum is equal to (1 + x) n . It is proved in the study of series that (15-10) is indeed valid if \x\ < 1. It follows that, if \x\ < 1, we may obtain the value of (1 + x) n as accurately as we please by taking sufficiently many terms of the series in (15-10). The expansion can readily be extended to (a + b) n when n is not a positive integer. In this case, the expression may be written as follows : (15-11) (a + 6)- = [a(l+J)]"=«-[l+|]'. Here x = - ? and the expansion of (a + b) n is valid if - < 1. a \a\ Example 15-14. Find the first five terms of the expansion of (1 + x)~ 3 if | x | < 1. Solution: By (15-10), (1 + *)-» = 1 + (■*■ 3)* + ( ~ 3) 2 f~ 4) s* (-3) (-4) (-5) '. (-3) (-4) (-5) (-8) + 3! x + 4| x + = 1 - 3x + 6x 2 - 10x 3 + 15x* + • • • . 272 Progressions Sec. 15-13 Example 15-15. Find -^L04. Solution: <i/TM = (1 + 0.04) ^ Hence, n = 1/3 and x = 0.04, and the ex- pansion is (1 + 0.04)"» = 1+| (0.04) + 2! (0.04)2 + 3 ^ 3 ^ *' (0.04)3+ .... The approximate value of the sum of this series is 1 + 0.013333 - 0.000177 + 0.000004 = 1.013160. We have here a case of an alternating series, that is, a series in which the terms are alternately positive and negative. It is proved in the study of series that, if in an alternating series each term is numerically less than the preceding term and lim a n = 0, the error introduced by using S n as the sum of the series is numerically less than the value of the first term omitted; that is, \S» - 8\ < |a»+i|. If in the present example we take for S the value S 3 = 1 + 0.013333 - 0.000177 = 1.013156, the error in so doing is less than the value of the fourth term 0.000004. Example 15-16. Find the first four terms of the expansion of — V8 -x 2 Solution: First we apply (15-11) to convert the given expression to a suitable form, as follows: (-r2\-l/3 1 / 3.2V -1/3 ^t) =10 -I) • Hence, by (15-10), if a; 2 < 8, _L_ = lfl | -1/3 / *\ , (-1/3) (-4/3) / g\» ■ViT^i* 2\ 1 V 8 J + 1-2 V 8/ | (-1/3) (-4/3) (-7/3) / g\» H \ L * 2 ' o V o / / 2 V ^ 24 ^ 288 ^ 20,736 ^ / EXERCISE 15-6 In each of the problems from 1 to 10, find the first four terms of a binomial expansion of the given expression. 1 « 1 1 + x 1 - x 3. VI + x. 4. VI - x. 5. (1 + x)-"*. 6. (i - x)-i/2. 7. -4— • 8. 7— r-vi • 9. 7 ,* x, ' 10. (1 + xy>*. x + 2/ (a + 1/) 2 (a; + y)* v y Find the approximate value of each of the following numbers by means of a binomial expansion, using four terms of the expression. 11. (1.02)". 12. (1.01) 13 . 13. (1.04) 8 . 14. (l.l) 10 . 15. (0.98) 8 . 16. 49 4 . 17. (0.99)«. 18. 51*. 19. (1.03) 1 ' 2 . 20. (0.97)" 2 . 16 Mathematical Induction 16-1. METHOD OF MATHEMATICAL INDUCTION When a certain type of formula or proposition has been verified in specific cases but is not known to be true in general, the method of mathematical induction is often found extremely valuable in determining its validity. Suppose that a statement involving a positive integer n is to be proved true for all values of n greater than or equal to a particular initial value. We begin by showing the result to be true for the first value of n. We then assume that k is some particular integral value of n for which the statement holds. With this assumption as a basis, we establish the validity of the statement in the next suc- ceeding case, namely, that in which n — k 4- 1. In other words, we prove that if the statement is true for any specific integral value of n, say n = k, then it is also true for the next larger value of n, namely, n = k + 1. Suppose for example, that 1 is the initial value of n. Then the second step establishes that if the statement is true for n = 1, it is also true for n = 1 + 1, or 2 ; if it is true for n = 2, it is also true f or n = 2 + 1, or 3 ; and so on. As a consequence of this, we conclude that the statement is true for all values of n greater than or equal to the initial value, here 1. A proof by mathematical induction, therefore, consists of two parts and a conclusion. Part 1. Verification that the statement is true for some initial, value of n, generally n = 1. (This initial value is the smallest value of n for which the statement is to be proved true.) Part 2. Proof that whenever the statement is true for some par- ticular value of n, say for n = k, then it is true for the next larger value of n, that is, for n = k + 1. 273 274 Mathematical Induction Sec. 16-1 Conclusion. If both parts of the proof have been given, then the statement is true for all positive integral values of n greater than or equal to the one for which the verification was made in Part 1. The reasoning process involved here, which consists in taking an initial integer and then repeatedly taking successors, can be exemplified in terms of climbing a ladder. Part 1 puts us on the bottom rung of a ladder. Part 2 shows us how to get from any rung we have reached to the next higher rung. The conclusion states that if we know how to get on the bottom rung of the ladder, and if we know how to get from any rung to the next higher one, then we can reach all rungs, and hence can climb the ladder. The following examples will illustrate the method. Example 16-1. If n is any positive integer, prove that (16-1) 1+-1-+J-- + ...+ 1 - n l-22-3^3-4 n(n + 1) n + 1 Solution: Part 1: The formula is true when n = 1, since 1 _ 1 11 1 • 2 "~ 1 + 1 ' ° r 2 2* Part 2: Let k represent any particular value of n for which (16-1) is true. Then (16-2) Y72 + 2T3 + 3T4 + * " * + k(k + 1) = F+T ' We now wish to prove that (16-1) is true also for the next larger value of n, namely, n = k + 1. The sum on the left in (16-1) when n = k + 1 can be obtained by adding its last term, which is ,, 1W , v > to both sides of (16-2). Hence, , v/c + l) {ic + Z) we have U-2 + 2-3 + 3-4i + *(* + !)/ ^ (& + !)(/ 1) (* + 2) k +■ ' k + 1 (* + 1) (* + 2) k 1 k 2 +2^ + 1 & + 1 Jfe + 1 But, since k + l , ( t + 1} ( t + 2 ) " (fc + l) (* + 2) " t + 2 " (* + 1) + 1 ' we obtain uo-a; i.2 i "2-3" t "3-4' t """" r "(* + l)(t+2)"(t+l)+l The members of (16-3) are the same as those of (16-1) when n = k -f 1. Hence, we have shown that (16-3) is true if (16-2) is true, in other words, (16-1) is true for n = k + 1 if it is true for n = k. Conclusion: We have shown by verification that (16-1) is true when n = 1. Therefore, since (16-1) is true for n = 1, it follows from Part 2 that (16-1) is true for every positive integer n. Sec. 16-2 Mathematical Induction 275 Example 16-2. Prove that (x - y) is a factor of (r» - y 11 ) if n is any positive integer. Solution: Part l:\ln = 1, then x n — y n = x — ?/, which is seen to have (x — y) as a factor. Part 0: Let A: be a specific value of n for which (x n — y n ) has (# - y) as a factor. We shall now prove that if {x — y) is a factor of (a;* — ?/*), it is also a factor of ( x *+i - y*+i). We have z*+i _ y* +1 = z* +l - xy* + zy* - y k+1 = x(x* - y k ) + */*(# ~ !/)• By assumption, (x — ?/) is a factor of (x k — y k ). Also, by inspection, (x — y) is a factor of t/*(x - y). Hence, (x - y) is a factor of the left member (x* +1 - y k+1 ). Therefore, if the conclusion is true for n = A*, it is also true for n = A; + 1. Conclusion: By Part 1 of the proof, (x - y) is a factor of (x n - ?/ n ) when n = 1. Therefore, by virtue of Part 2, the desired conclusion is true for any positive integer n. 16-2. PROOF OF THE BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENTS We shall now prove that the binomial formula, (16-4) (a + &)» =a n + na n ~ l b + n{jl ~ ^ a n ~ 2 b 2 n(n-l)---(n-r + 2) ^^ (r - 1)! , n(n — 1) • • • (n — r + 1) , . -| — >. ^ p £ a n r b r + • • • + o w > r! is true for every positive integral value of n. Proof. Part 1. When n = 1, each side of (16-4) becomes a + &. Hence, (16-4) is true for n = 1. Par£ #. Let /c be any specific value of n for which (16-4) is true. Then we have (a + b) k = a* + ka k ~ l b + k<<k ~ ^ a k ~ 2 b 2 r! Multiplying each member of this equation by (a + b), we obtain ( a + 6 )*+i = a *+i + fca *& + . . . + * ( * = 1} - \ (fc = r + 1} a*-'* 1 **' r\ + • • • + ab k + a*6 + • • • + W-l)...(fe-r + 2) ^ r+1&r + _ + fct+1 (r - 1) ! 276 Mathematical Induction See. 16-2 Hence, by combining terms, we get (o + b) k+1 = a* +1 + (Jfc + l)a*6 + • • • (ib + l) t ...(fc-r + 2) ^^ ftt+1 r! In this result the sum of the coefficients of a k - r+1 b r is obtained as follows : k( k - 1) - --(ft -^r + 2) (fe - r + 1) fc(Jfc - 1) . . . (Jfe - r + 2) r! -P^±! +.].[: (r-1)! fc(Jb - 1) ; ; ; (fc - f + 2)' r J l (r-1)! J _ (fc + 1) (fc) (fc - 1) • » ■ (Jb - r + 2) r! We note that the value of (a + b) k+1 , obtained as the product of (a + &)* and (a + 6), is exactly the same as the expansion which would be obtained from (16-4) with n = fc + l. Hence, we have shown that if the binomial formula holds for n = k, it must hold for n = fc + 1. Conclusion. The binomial formula was seen to be true for n = 1. Therefore, by virtue of Part 2, we may conclude that it is true for every positive integer n. EXERCISE 16-1 Prove by mathematical induction that each of the statements in Problems 1 to 20 is true for all positive integral values of n. , n(n + 1) 1. 1 + 2 + 3 + • 2. 1 + 3 + 5 + • 3. 2 + 4 + 6 + • 4. 1 + 3 + 6 + • 5. 3 + 6 + 9 + • 6. 4 + 8 + 12 + 7. 1 + 4 + 7 + • 8 i+i + i + . 8 * 2 + 4 + § + 9. 1 • 2 + 2 • 3 + 3 10. 1* + 2* + 3* + + n =■ 4- (2n - 1) = n 2 . + 2n = n(n + 1). n(h + 1) _ n(n + 1) (n + 2) 2 ~~ 6 + + 8n =§2fiLtl). • +4n =2n(n + 1). + (3n -2)=^L=J). n . _ «(n + 1) (an + 1) , Sec. 16-2 Mathematical Induction 277 11.2»+4»+6»+--- + (2n)'= 2w(w + 1 > (2w + 1) . o 12 . P+23+33 + . .. +n . = [2fi^tiiJ. 13. I 3 + 3 3 + 5 3 + • • • + (2n - l) 3 = n 2 (2r* 2 - 1). 14. 2 3 + 4 3 + 6 3 + • • • + (2n) 3 = 2n 2 (n + I) 2 . 15. 2 + 2 2 + 2 3 + • • • +2« = 2(2» - 1). 16. 3 + 3 2 + 3 3 + • • • + 3» = | (3» - 1). 17. 4 + 4 2 + 4 3 + • • • + 4» = | (4» -,1). 18. I 2 + 3 2 + 5* + 7 2 + • • • + (2n - | 2 = n(2n - 1) gn + 1) . 19 ±, 1 , 1 , . 1 _Vn 1-3 ' 3 • 5 ^ 5 • 7 ^ ^ (2n - 1) (2n + 1) ~ 2n + 1 20 1 i 1 i 1 i , * = n(n+3) . 1 • 2 • 3 ^ 2 • 3 • 4 **" 3 • 4 • 5 ^ "*" n(n + 1) (n + 2) 4(n + 1) (n + 2) 21. Prove that # 2n - y 2n is divisible by a: -f- y for every positive integer n. 22. Prove that a; 2 * 1 - 1 + y 2n ~ l is divisible by x + 2/, for every positive integer n. 23. By using mathematical induction, prove the formula for the sum of an arith- metic progression. 24. By using mathematical induction, prove the formula for the sum of a geometric progression. 17 Permutations, Combinations, and Probability 17-1. FUNDAMENTAL PRINCIPLE We begin the study of permutations and combinations by con- sidering the following principle, which is fundamental for the entire subject. Fundamental Principle. If one thing can be done in a ways, and if, for each such way, a second thing can be done in b ways, then the two together can be done in a • b ways. To understand why the principle is true, note that for each of the a ways of doing the first thing, there are b ways of doing the second ; hence, both the first and the second things taken together can be done in a • b ways. The following examples will illustrate the reasoning upon which the principle is based, as well as an obvious extension of the prin- ciple to the case when more than two things are to be done. Example 17-1. In how many ways can two officers, a chairman and a secretary, be selected from a committee of five men? Solution: By the fundamental principle, the problem is equivalent to determining the number of ways in which the two things can be done together. The first position can be filled in 5 ways; that is, there are a = 5 ways of selecting a chairman. For each of these possible selections, there are b = 4 ways of filling the position of secretary from the remaining men. Hence, the number of ways of selecting a chairman and secretary is a • b = 5 • 4 = 20. Example 17-2. How many three-digit numbers can be formed from the ten digits 0, 1, 2, • • • , 9, if a) repetitions of digits are not permitted; 6) repetitions are permitted? Solution: a) Here we have three things to do or places to fill. The hundreds place can be filled in 9 ways, since must be excluded from this place. The tens 278 Sec. 17-2 Permutations, Combinations, and Probability 279 place can then be filled in 9 ways from any of the remaining 9 digits. Finally, the units place can be filled from any of the remaining 8 digits. There are, therefore, 9.9.8 = 648 three-digit numbers in which no two digits are alike. b) If repetitions are permitted, there are 9 ways to fill the hundreds place and 10 ways to fill each of the tens and units places. Hence, there are 9 • 10 • 10 = 900 three-digit numbers. EXERCISE 17-1 1. Nine persons apply for each of two vacant apartments. In how many possible ways can both apartments be rented? 2. A large room has eight doors. In how many ways can a person enter the room by one door and leave by a different door? 3. If three dice are thrown, in how many ways can they fall? 4. There are eight men and six women in a club. In how many ways can two officers be selected so that one is a man and one is a woman? 5. How many possible four-digit numbers are there in a telephone exchange which uses only four-digit numbers and excludes 0000? 6. How many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6, 7, 8, 9? How many of these are larger than 700,000? 7. How many numbers of six different digits can be formed from the digits 2, 3, 4, 5, 6, 7, 8, 9? How many of these are larger than 700,000? 8. A woman has seven guests at a party. If she chooses her seat first, in how many ways can she seat her guests? 17-2. PERMUTATIONS An ordered arrangement of all or any part of a set of things is called a permutation. Specifically, suppose we have n distinct things and wish to select r of these to be arranged in a definite order. Each such ordered arrangement is called a permutation of n things r at a time. The number of all such permutations is denoted by n P r . Thus, 5 P 2 is read "the number of permutations of 5 different things 2 at a time." In Example 17-1, the possible number of ways of selecting a chairman and secretary from a committee of five men was found by means of the fundamental principle to be 5 • 4 = 20. This is pre- cisely equal to 5 P 2 , since it is the number of ways in which two men can be chosen from among the given five men and arranged in the two offices. The number of permutations of n things rata time is given by the formula (17-1) n P r = n(n - 1) (n - 2) • • • (n - r + 1). The truth of (17-1) is readily shown as follows. The first of the r places can be filled in n ways. Then the second can be filled in (n — 1) ways, the third in (n — 2) ways, and so on. In general, the 280 Permutations, Combinations, and Probability Sec. 17-2 number of ways of filling each place is n minus the number of places already filled. Therefore, when the rth object is chosen, (r — 1) places have already been filled, and the rth place can then be filled inn- (r — 1) , or n - r + 1, ways. In particular, if r = n, the last factor becomes n — n + 1 = 1. We then have (17-2) n P n = n(n - 1) (n - 2) • • • 1 = n!. This formula gives the number of permutations of n different things taken all at a time. If both the numerator and the (understood unit) denominator of (17-1) are multiplied by (n — r) !, we obtain the following alter- nate formula : (17-3) P - n ( n ~ *) ( n ~ 2 ) ' ' * ( n ~ r + *) ' ( n ~ r ) ! = n! ^ ' n r (n — r)\ (n — r)\ Here it is agreed that by definition ! = 1. Hence, (17-3) holds for r = n. For example, by (17-2), 8 P 6 = n(n - 1) (n - 2) • • • (n - r + 1) = 8 • 7 • 6 • 5 • 4 = 6720. Using (17-3), we have also j, n! 8! 8-7-6-5-4-3-2-1 A7on •ft = (S-zt^i = si = F2^ = 672a 17-3. PERMUTATIONS OF n THINGS NOT ALL DIFFERENT Suppose it is required to find the number of indistinguishable distinct permutations of n things all at a time, if w x things are regarded as indistinguishable, n^ other things are regarded as indis- tinguishable, and so on. Let us consider, for example, the number of permutations of the letters a, a, a, b taken four at a time. For convenience, indistinguishable objects are given the same notation. Denote by P the desired number of distinct permutations. Evi- dently, P is less than 4P4, which is the number of permutations that could be effected if all the letters were distinguishable. For in any one of the P permutations, say (a b a a) , any rearrangement of the a's among themselves would not change the permutation. If, how- ever, we assigned subscripts to a in this permutation, as in (a 2 6 a® a 3 ), we could permute these three distinct letters among themselves in 3 ! ways. This can be done for each of the P permu- tations of the letters a, a, a, b. We would then obtain P • 3 ! per- mutations of the four distinct letters a u ch, a 3 , b taken four at a time. There are therefore 4 ! permutations altogether. Hence, 4f 4! P. 8! =4!, or P = |j = %l 3! 3!1! Sec. 17-4 Permutations, Combinations, and Probability 281 In general, the number of distinct permutations of n things taken all at a time, if n x things are alike, n 2 other things are alike, n 3 other things are also alike, and so on, equals n! (17-4) P = 17-4. COMBINATIONS n\\ ri2\ n$\ A combination is a set of all or any part of a collection of objects without regard to the order of the objects in the set. We use the symbol n C r to represent the total number of all combinations of n different things taken r at a time. The different sets of the four letters a, b, c, d taken three at a time, without reference to the order in which the letters are arranged, are (abc), (abd), (acd), (bed). From each of these four combinations, we can form 3 !, or 6, different permutations of the four letters taken three at a time. For example, from the combina- tion (abc) we can form the distinct permutations (abc), (acb), (bac), (bca), (cab), (cba). Hence, each of the four combinations contributes 3 P 3 = 3 ! permutations to the total number of permu- tations. Thus, there are 4 C 3 combinations of the four letters if we disregard order, and there are 3! ordered arrangements or per- mutations of each combination. Hence, we have 4 P 3 = 3 ! • 4 C 3 , or 4 C 3 = ~y = ' ' ' = 4 combinations. This is the same as the o! 1 • 2 • o number we obtained at the beginning of the paragraph. In general, if we divide the total number of permutations of n things rata time, or n P rp by the number of permutations, or r!, contributed by each combination, we obtain the total number of combinations. Symbolically, we have or (17-5) n C r = „, P If we note that r ! = r P r , then we can write the following inter- esting relationship : o — n r . nW — p Replacing n P r by its equivalent expression from (17-1) or (17-3) we have (17-6) nC r 7{ r!(n-r-r)! 282 Permutations, Combinations, and Probability Sec. 17-4 — r) in If r is replaced by (ft — r) in (17-6) , we obtain n! (n — r)! r! Hence, Note. Having agreed that ! = 1, we can supply (17-1) or (17-3) to permutations or combinations of n things zero at a time. Con- sidering (17-5), we have n! 1 (17 ~ 8) nCo = 0!(ft-0)! = 0! = L This result agrees with the intuitive conclusion that there is only one such "empty" combination. Similarly, n P = 1. 17-5. BINOMIAL COEFFICIENTS By referring to the development of the binomial formula in Sec- tion 4-6, we note that the expansion of (a + b) n involves the prod- uct (a + b) (a + b) (a + b) • • • taken without regard to order. This fact suggests the use of combinations in the coefficients of the expansion. We said in Section 4-7 that the coefficient of the term involving a n ~ r b r is — - - - - . ~~ r • This is precisely the formula for r! the combination of n things rata time, or n C r , obtained in Section 17-4. The binomial formula may therefore be written as follows : (17-9) (a+b)» = n Coa«+ n C 1 a n - l b+ n C 2 a n - 2 b 2 +. • .+ n C r a-^+. . -+ n CJ>\ For example, the expansion of (a + b) 4 takes the following form: 4 C a 4 + 4 Cia 3 5 + 4 C 2 a 2 6 2 + *Csab 3 + 4 C 4 6 4 . This reduces to a 4 + 4a 3 b + da 2 b 2 + 4ab 3 + b 4 . If we let a = b = 1 in the expansion for (a + b) n in (17-9), we obtain (1 + 1)" = n C + nCl + n C 2 + ' • • + nC n . Since n C = 1, n Ci + n C 2 + • • • + n C n = 2* - 1. Thus, the total number o;f combinations of n things taken succes- sively 1, 2, • - - , n at a time is 2* — 1. EXERCISE 17-2 1. Evaluate 5P2, 7P3, 12^6, 21P4, 100F3. 2. Evaluate 10C4, 11C3, iooCsj 21C5, iooCos* Sec. 17-6 Permutations, Combinations, and Probability 283 3. Form all possible distinct permutations of the letters of the word theory, a. How many are there? b. How many begin and end with a vowel? c. How many begin or end with a vowel? 4. How many distinct permutations are there of the five letters a, b, c, d, e taken three at a time? Write them out. 5. In how many different ways can a dime, a quarter, and a half-dollar be dis- tributed among five boys? 6. How many different sums can be formed with a penny, a nickel, a dime, and a quarter? 7. Expand (a -f 6) 8 , using combination symbols. 8. From the six digits 1, 2, 3, 4, 5, 6, form all permutations taken five at a time. a. How many are formed? b. How many begin with 4? c. In how many does the digit 3 not appear? 9. How many distinct permutations can be made from the letters of the word probability taken all at a time? 10. How many different combinations are there of 4 identical nickels ; 5 identical dimes, and 6 identical quarters? 11. How many different straight lines are determined by twelve points, no three of which are in a straight line? 12. In how many different ways can a student select seven questions out of ten on a test? 13. How many different weights can be formed with six objects weighing 1, 2, 4, 8, 16, and 32 pounds, respectively? 14. In how many different ways can signals be made with seven different flags, where a signal is a set of one or more flags arranged in a specific order? 15. In how many different ways can the 52 cards of a bridge deck be dealt among four players? 17-6. MATHEMATICAL PROBABILITY If, in a given trial, an event can happen in h different ways and can fail to happen in / ways, and if all the h + f ways are equally likely, then the probability p that the event will happen in this trial is (17-10) p= 1 ± ? . The probability q that the event will fail to happen is (17-11) , = r ^_. Note that O^p^l, ^q^l f and p + q = 1. Two illustrations of probability follow. If a bag contains 3 green marbles and 4 yellow marbles, all exactly alike except for color, the probability of drawing a single green marble is 3 3 V 3 + 4 7 # 284 Permutations, Combinations, and Probability Sec. 17-6 A die can fall in six ways. The probability of getting 5 or more 2 1 ith one throw of a die is - > or - > since th o o either a 5 or a 6, for the event to happen. 2 1 with one throw of a die is - > or - > since there are two possible ways, o o 17-7. MOST PROBABLE NUMBER AND MATHEMATICAL EXPECTATION Let p be the probability of occurrence of some event. Further- more, suppose that n trials of the event are made, of which h are successful. Then - is called the relative frequency of success for n the trials which occurred. It is not to be expected that - = p. It is shown in more advanced treatments of probability, however, that if n is large, it is very likely that the relative frequency is approxi- mately equal to p. Also, the larger we take n, the more likely it is that - approximates p closely. Moreover, it can be shown that the n most probable or expected number of occurrences of the event for n trials is np. For example, when a coin is tossed, the probability of getting a head is 1/2. In 1,000 trials the expected number of heads is there- fore 500. This does not mean, however, that if the first 100 trials result in 75 heads and 25 tails, we should expect 25 heads and 75 tails in the next 100 trials. Actually, since one toss of the coin does not affect the next one, we should expect about 50 heads and 50 tails in the next 100 trials. Moreover, we may expect about 450 heads and 450 tails in the next 900 trials. If p is the probability of winning a certain amount of money in case a certain event occurs and m is the amount of money to be won, the mathematical expectation is defined to be pm. For instance, if a person can win $12 provided he throws an ace with a die, his expec- tation is - ($12) = $2. Hence, $2 is the fair amount he should be willing to pay to make the trial. 17-8. STATISTICAL, OR EMPIRICAL, PROBABILITY It is frequently impossible to have sufficient knowledge before- hand of all the conditions that might cause an event to happen or fail to happen. In such a case, however, it may be possible to deter- mine the relative frequency of the occurrence of the event from a large number of trials. Thus, if an event has been observed to happen h times in n trials, and n is a large number, then until addi- Sec. 17—9 Permutations, Combinations, and Probability Hl$ tional knowledge is available, we define the statistical probability, or empirical probability, to be (17-11) p=l> where - is the relative frequency. n Example 17-3. A molding machine turns out 12 parts per minute. Inspection experience has shown that there are 20 defective parts per hour. What is the probability that a single part, picked at random, will be defective? In a run of 10,000 parts, how many defectives should be expected? Solution: The parts are produced at the rate of 720 per hour, and 20 of them are defective. Hence, the probability that a single part selected at random will be defective is =xr = ™ • In a run of 10,000 parts, we should expect ^ (10,000), or approximately 278, defective parts. 17-9. MUTUALLY EXCLUSIVE EVENTS Two or more events are mutually exclusive if not more than one of them can happen in a given trial. The following theorem may be stated. Theorem. The probability that some one of a set of mutually exclusive events will happen in a given trial is the sum of the indi- vidual-event probabilities. Proof. Consider, for simplicity, a set of two mutually exclusive events. Suppose that the first can happen in h x ways and the second can happen in h 2 ways, and let n be the total number of ways in which the two events can happen or fail to happen. Then p\ = — lb and p2 = — are the corresponding probabilities of the two events. Tt Since the n events are mutually exclusive, the h x ways are differ- ent from the h 2 ways, and the number of ways that either the first or the second event can happen is therefore hi + th. Hence, the probability p that either the one event or the other will happen is ,+ *, 10N hi + hi h\ h 2 (17-12) p= ___ = _ + _ = pi -r p2 . For example, suppose that a bag contains 2 green marbles, 3 yel- low marbles and 5 brown marbles. If a marble is drawn at random, 2 the probability that it is green is ^ > and the probability that it is 3 yellow is — • Hence if a marble is drawn, the probability that it is .., „ .2,3 51 either green or yellow is — + ^ > or ^ > or ^ • 286 Permutations, Combinations, and Probability Sec. 17-10 17-10. DEPENDENT AND INDEPENDENT EVENTS In case two or more events are not mutually exclusive, they are dependent if the occurrence of any one affects the occurrence of the others, and they are independent if the occurrence of one does not affect the occurrence of the others. For example, if a card is drawn from a deck of 52 cards and the card is not replaced before a second is drawn, then the second drawing is dependent on the outcome of the first. If, however, the first card is replaced, then the second drawing is independent of the first. In the latter case the two drawings are equivalent to simultaneous drawings from two decks. We shall now state and prove the following theorem relating to dependent and independent events. Theorem. The probability that two dependent or independent events will occur (successively if dependent; successively or simul- taneously if independent) is the product of their individual probabilities. Proof. Suppose that the first event can happen in h x out of a total of n x different ways, and that the second event can happen in hz out of n*> different ways. Then it follows, by the fundamental principle in Section 17-1, that the two events can happen together in hxh 2 ways out of a total of n x n 2 different ways. Therefore, the probability that both events will happen is (17-13) p = = — • — = p\P2> U\U2 U\ n 2 Example 17-4. Two cards are drawn from a deck containing 52 cards. Find the probability that both cards are aces when the first card is not replaced before the second is drawn. Solution: We shall begin with a listing of the following useful probabilities: 4 1) The probability of drawing an ace from a deck of 52 cards is ~ • uZ 2) If the first card is an ace and it is not replaced, the probability of drawing another ace is -=^ • ol 3) If the first card is not an ace and is not replaced, the probability of the second being an ace is — • ol 4 4) If the first card is replaced, the probability of the second being an ace is ^ • This drawing is entirely independent of the first drawing. Consider, now, the given problem. The probability that the first card is an ace is 4 1 P l = ^9 = 13 " ^ ^ e ^ rs ^ carc * drawn * s an ace > ^ nen the probability that the Sec. 1 7-1 1 Permutations, Combinations, and Probability 287 3 1 second card drawn is an ace is P2 = ~y = j~ • Hence, the probability that both •ii u • 111 will be aces is pip 2 = 13 ' 17 = 221 * 17-11. REPEATED TRIALS Theorem. If p is the probability that an event will happen in any trial, and q = 1 — p is the probability that it will fail, then the prob- ability that it will happen exactly r times out of n trials is (17-14) nCrP'q*-* = f{ fr]_ f)[ PT^ Proof. The happening of the event in exactly r trials and its fail- ure in the remaining n - r times are independent events. Hence, the probability, by the theorem in Section 17-10, is p r q n ~ r . But these r trials can be chosen from the n trials in n C r ways. Since these ways are mutually exclusive, the total probability is n C r p r q n ~ r * Note that this expression is the (r + l)th term of the binomial formula for (q + p) n , since (q + p)» = q" + nCiq^p + n C 2 q n ' 2 p 2 + h nC r q n - r p r + • • • + p n . The successive terms of this expansion give the probabilities that the event will happen exactly 0, 1, 2, • • , r, • • • n times in n trials. An event will happen at least r times in a given number of n trials if it happens n, n — 1, • • • , r + 1, or r times. Since these events are mutually exclusive, the probability that an event will happen at le&st r times is given by the sum P« + n C n _l qp"" 1 + --'+nC r q n ~ r p\ Example 17-5. What is the probability of tossing an ace exactly three times in four trials with one die? Solution: Since the probability of tossing an ace in one trial is ^ and the prob- 5 ability of failure is -r 1 we may substitute in the term n Cr5 rn " r P r of the binomial formula. The result is 4C7s (DXI)^ 4 (I) (lie) = sli* Example 17-6. What is the probability of tossing an ace at least twice in four trials with one die? Solution: The event will happen at least twice if it happens 4, 3, or 2 times. Hence, the probability is given by the following sum : •Mi)'W!)a)'wf)'a)'=ir 288 Permutations, Combinations, and Probability Sec. 17-1 1 EXERCISE 17-3 1. A certain event can happen in four ways and can fail to happen in six ways. What is the probability that it will happen? If $60 can be won on the event, what is the mathematical expectation? 2. A box contains 5 white balls, 4 red balls, and 13 black balls, a. If one ball is drawn out, what is the probability that it is red? b. What is the probability that it is white or red? 3. a. If one die is thrown, what is the probability that a "1" or a "2" will turn up? b. What is the probability that a "3" or larger number will turn up? 4. When a coin was tossed 100 times, 80 heads and 20 tails turned up. If the tossing were continued until 200 tosses had been made, what would be the most probable number of tails in the second 100 tosses? 5. A bag contains five $1 bills, ten $5 bills, and twenty $10 bills. If one bill is drawn, what is the mathematical expectation? 6. What is the probability of throwing a "7" or an "11" on one throw of two dice? 7. An automobile owner carries $1,000 theft insurance on his car. If, during the past year, 237 out of 97,864 automobiles registered in his area were stolen, what is the mathematical value of the policy? 8. In a city of 77,000 families, a careful sample of 800 families showed that 120 of the sample families owned their own homes, a. What is the probability that a family selected at random in the city owned its home? b. What is the expected number of families in the city who own their own homes? 9. In a certain city 28,600 persons voted for one candidate for an office, and 23,100 voted for his opponent. What is the probability that a voter chosen at random voted for the winner? 10. A bag contains 5 red balls and 9 black balls. If two balls are drawn in suc- cession, and the first is not replaced, find the probability that the first is red and the second is black. 11. In a baseball tournament the probability that team A will win is = > and the 1 probability that team B will win is ^ • Find the probability that one of these two teams will win. 2 12. The probability that team A will reach the finals of a tournament is » > 1 7 and the probability that it will win the finals is - • Find the probability that team A will win the tournament. 13. Find the probability of throwing three successive fours on a pair of dice. 14. The probability of A winning a game when he plays it is j • He is scheduled to play four times, a. Find the probability that he will win exactly three times. b. Find the probability that he will win at least three times. 15. Three dice are tossed, a. Find the probability that exactly two threes will turn up. b. What is the probability that at most two threes will turn up? 18 Solution of the General Triangle 18-1. CLASSES OF PROBLEMS There are certain relationships among the lengths of the sides and the trigonometric functions of the angles of every triangle. If one side and any two other parts of a triangle are given, the remaining parts can be determined; that is, the triangle can be solved. The three given parts may comprise any one of the follow- ing four combinations : Case I. One side and two angles. Case II. Two sides and the angle opposite one of them. Case IIL Two sides and the included angle. Case IV. Three sides. In this chapter, we shall discuss methods for treating these four cases. For convenience, we shall let ABC denote any triangle whose angles are A, B, and C; and we shall let a, b, and c represent the lengths of the corresponding opposite sides. 18-2. THE LAW OF SINES Law of Sines. Let ABC be any triangle lettered in the conven- tional manner. Then the following relationship between the sides and the sines of the angles may be written : (18-1) « = * = «. sin A sin B sin C This relationship is commonly called the law of sines. c B A Fig. 18-1. 289 290 Solution of the General Triangle Sec. 1 8-2 Proof. We first note that all angles may be acute, as in Fig. 18-1 (a), or one angle, say B, may be obtuse, as in Fig. 18-1(6). (The case where B = 90° entails no difficulties and will therefore be omitted.) In each diagram, let h denote the altitude from the vertex C to the side AB. Then, in either case, sin A = r and h sin B = - • Dividing the first equation by the second, we have a sin A a a b or sin B b sin A sin B In a similar way, by drawing the altitude from the vertex A to the side BC, we get b c sin B ~~ sin C The equations thus obtained may be combined to give the law of sines sin A sin B sin C Note. The law of sines is well adapted to the use of logarithms because it involves only multiplications and divisions. Since any pair of ratios in the law of sines involves two angles and the sides opposite, it may be used in the solution of problems in Cases I and II. As noted above, the law of sines also applies in the special case where ABC is a right triangle. In this case, one of the angles is 90°, and the sine of that angle is 1. 18-3. SOLUTION OF CASE I BY THE LAW OF SINES: GIVEN ONE SIDE AND TWO ANGLES When one side and any two angles of a triangle are known, the third angle can be found from the relation A + B + C = 180°, and each of the required sides is uniquely determined. These sides may be found by the law of sines. Example 18-1. In a triangle ABC, A = 38°14', B = 67°20', c = 329. Solve the triangle. Solution: The values of the given parts are indicated in Fig. 18-2. In this case, c C = 180° - (38°14' + 67°20') = 74°26'. To find a, we use the relationship a a 329 sin 38°14 / sin 74°26' _____ Therefore, c-329 " 329 sin 38°14 / Fig. 18-2. a "" sin 74 26' ' Sec. 18-4 Solution of the General Triangle 291 and log a = log 329 + log sin 38°14' - log sin 74°26\ The indicated operations may be performed as follows: log 329 = log sin 38°14' = log sin 74°26 / = log a = 2.5172 9.7916 - 10 12.3088 - 10 9.9838 - 10 2.3250 a =211. To determine b, we use the relationship 329 Hence, and sin 67°20' ~ sin 74°26' 329 sin 67°20' 6 =: sin 74°26' log b = log 329 + log sin 67°20' - log sin 74°26'. The work follows: log 329 = 2.5172 log sin 67°20' = 9.9651 - 10 log sin 74°26' = log b = 12.4823 - 10 9.9838 - 10 2.4985 b = 315. As shown in Fig. 18-1 (a), c = b cos A + a cos B. This relationship may be used as a check. Thus, c = 315 cos 38°14' + 212 cos 67°20' = (315) (0.7855) + (212) (0.3854) = 329.1, which agrees satisfactorily with the given value of c. 18-4. SOLUTION OF CASE II BY THE LAW OF SINES GIVEN TWO SIDES AND THE ANGLE OPPOSITE ONE OF THEM Case II is called the ambiguous case, because the data may be such that two, one, or no triangles are determined. The number of solutions when a, b, and A are given is indicated by the accompany- ing table. Table of Possible Solutions (18-2) (18-3) (18-4) (18-5) A acute a = b sin A a <b sin A b sin A < a <b a^b One right triangle No triangle Two triangles One triangle (18-6) (18-7) A obtuse a^b a>b No triangle One triangle 292 Solution of the General Triangle Sec. 18-4 We shall use Fig. 18-3 to illustrate in turn the different possi- bilities considered in the table. If the angle A and the sides a and b are given, we first construct the angle A with the initial ray AX and the terminal ray AR. Next, we lay off the distance AC = 6 along the terminal side. Then, with C as the center and the length of the side a as the radius, we describe an arc. We mark the point or points in which this arc intersects the initial ray AX of the angle A. Fig. 18-3. Figure 18-3 (a) corresponds to (18-2) in the table. Since BC = a = b sin A, this segment is the altitude of the triangle drawn from the vertex C. Hence, the arc with radius a is tangent to the initial side at B, and the triangle is a right triangle. Figure 18-3(6) corresponds to (18-3) in the table. Since a < b sin A, the side a is too short to intersect AX, and there is no triangle. In Fig. 18-3 (c), a < b and b sin A < a, as stated in (18-4) in the table, and the arc will intersect AX in two points marked B and B'. Therefore, two solutions exist. The angle 2?' in the triangle AB'C is the supplement of the angle B in the triangle ABC. Figure 18-3 (d) represents the case in which the side a is longer than the side 6, as stated in (18-5) in the table. Hence, there is only one point B in which the arc with radius a intersects the initial ray AX of the angle A. There is only one solution. Sec. 18-4 Solution of the General Triangle 293 In Fig. 18-3 (e), the angle A is obtuse. Since a < b, as stated in (18-6), the radius a is too short to intersect the initial ray AX, and no triangle exists. Finally, in Fig. 18-3 (/) , A is obtuse and a > b, as stated in (18-7). Here the arc can intersect the initial ray AX in only one point. There is, in this case, only one triangle. The following examples will illustrate some of the possibilities. Example 18-2. In a triangle ABC, A = 36°15', a = 9.8, b = 12.4. Solve the triangle. Solution: Draw Fig. 18-4 approximately to scale, showing the given parts. After angle A and side b have been drawn, an arc is described with C as center and a as radius. The arc intersects the side AX in two points, B\ and B 2 , and we apparently have two possible triangles, AB X C and AB 2 C. Fig. 18-4. To find sin B, we use Then sin B sin A sin B = b a 12.4 sin 36°15' 9.8 and log sin B = log 12.4 + log sin 36°15' - log 9.8 The logarithmic work follows: log 12.4 = 1.0934 log sin 36°15 ; = 9.7718 - 10 10.8652 - 10 log 9.8 = 0.9912 log sin B = 9.8740 - 10 B = 48°26'. There are two solutions, since 6 sin A < a < b. The left inequality follows from the fact that log sin B = log ^*A < o, and so ^4 < 1. If we let B x = 48°26', 294 Sblution of the General Triangle Sec. 18-4 then B 2 = 180° - 48°26' = 131°34' leads to another solution. Thus, B x = 48°26', Ci = 180° - (36°15' + 48°26') = 95°19', and B 2 = m^', C 2 = 180° - (36°15' + 131°34') = 12°11'. To find Ci, we use the law of sines again. Thus, 9.8 sin 95°19' Ci = Similarly, we have c 2 = ■ sin 36°15' 9.8 sin 12°11' 16.5. = 3.5. sin 36°15' As a partial check, the equation Ci = b cos A + a cos Bi may be used. Thus, 12.4 cos 36°15' + 9.8 cos 48°26' = 12.4 (0.8064) + 9.8 (0.6635) = 16.5. This result is the same as the value previously calculated. The same method may be applied to check c 2 . Example 18-3. Given A = 56°30 , a = 13.0, b = 10.7, solve the triangle ABC. Solution: Here we clearly have only one solution, since a > b. This can be seen geometrically if we draw Fig. 18-5 approximately to scale and show the given parts. Since a is greater than b, an arc with C as center intersects AX on opposite sides of A. Obviously there is only one triangle, ABiC, containing the angle A. To find J5i, we use the relationship sin Bt sin 56°30' a - 13.0 10.7 13 whence x log sin Bi = log 10.7 This gives Fig. 18-5. Then C = 180° - (56°30' + 43°20') = 80°10'. To find c, we use the law of sines and obtain c 13 + log sin 56°30' - log 13. B l = 43 o 20'. sin 80°10' sin 56°30' This gives c = 15.4. Example 18-4. Given A = 67°40', a = 16.0, b = 17.3, solve the triangle. C Solution: The given parts are shown in Fig. 18-6. By the law of sines, we have sinB _ sin 67°40 / 17.3 " 16 Therefore, log sin B = log 17.3 -f log sin 67°40' - log 16 = (1.2380) + (9.9661 - 10) - (1.2041) = 0. Hence, B = 90°. The solution may be completed by applying the theory of right triangles. Only one solution exists. Sec. 18-4 Solution of the General Triangle 295 Example 18-5. Given A = 47°23', a = 230, b = 720, solve the triangle. Solution: From Fig. 18-7, it appears that no triangle is possible. The following work verifies this fact. To find B, use the relationship sin B sin A obtaining sin B = b a 720 sin 47°23' 230 a -230 The logarithmic work follows: log 720 = 2.8573 log sin 47°23' = 9.8668 - 10 12.7241 - 10 log 230 = 2.3617 Fig. 18-7. log sin B = 10.3624 - 10 = 0.3624. Since log sin B > 0, sin B would have to be greater than 1, which is impossible. Therefore, there is no solution. EXERCISE 18-1 In each of the problems from 1 to 12, solve the given triangle by the law of sines. 1. a = 12.30, A = 36°25', B = 44°37'. 2. b = 12.18, A = 47°33', B = 67°51'. 3. c = 461.3, B = 67°19', C = 23°14'. 4. b = 0.6384, B = 39°39', C = 87°16'. 5. a = 6.714, A = 37°53', C = 136°36'. 6. c = 7832, A = 68°39 ; , B = 43°58 ; . 7. a = 21.23, c = 64.21, C = 62°31'. 8. b = 0.8146, c = 31.63, B = 11°10 / . 9. a = 987.4, b = 503.6, A = 54°13'. 10. a = 0.003862, c = 0.0008157, A = 26 13 ; . 11. b = 1.386, c = 2.451, 5 = 83°19 ; . 12. b = 4.395, c = 9.806, C = 37°46'. 13. A surveyor wishes to find the distance across a stream from point A to point B. He finds that the distance from A to a point C on the same side of the stream is 687.4 feet, and angles BAC and BCA are 49°53' and &8°16 ; , respectively. Find the distance AB. 14. A surveyor was running a line due west when he reached a swamp. From the edge of the swamp he ran a line S 63° W for 2500 feet, and from this point he ran a line N 27 23' W. How far had he gone on this line when he reached his original line produced? How far was it across the swamp? 15. A building 63.7 feet high stands on the top of a hill. From a point at the foot of the hill the angles of elevation to the top and bottom of the building are 42°16' and 38°31', respectively. Find the height of the hill. 296 Solution of the General Triangle Sec. 18-4 16. From a certain point on the ground the angle of elevation to the top of a building is 46° 17'. From a point on the ground 83 feet nearer the building the angle of elevation is 68°23'. Assuming that the ground is level, find the height of the building. 17. One side and a diagonal of a parallelogram are 14.63 inches and 21.4 inches, respectively. The angle between the diagonals and opposite the given side is 116°23'. Find the length of the other diagonal. 18. It is necessary to measure the distance between two artillery pieces A and B. The angle of depression from an observation point C to gun A is 24°47'. Sound travels at the rate of 1 140 feet per second, and the sounds from guns A and B reach C in 2.3 and 1.7 seconds, respectively. Find the distance AB } assuming that points A } B ) and C lie in the same plane. 19. A body is acted on by two forces, Fi = 2643 pounds and F 2 = 2341 pounds. The resultant F 3 lies on a line making an angle of 46°33' with F x . Find F 3 and the angle between the lines of action of Fi and F 2 . (The resultant of two forces is their vector sum.) 20. A buoy, located at a point B, is 6 miles from a point A at one end of an island and 10 miles from a point C at the other end of the island. If the angle BAC is 132°16', find the distance between the points A and C on the island. 18-5. THE LAW OF COSINES Law of Cosines* Let ABC be any triangle. Then (18-8) a 2 = b 2 + c 2 - 2bc cos A, (18-9) b 2 = a 2 + c 2 - 2ac cos B, (18-10) c 2 = a 2 + b 2 - 2ab cos C. These relationships constitute the law of cosines. < n V h >^o A ^ X D c-x V- - c (a) B Fig. 18-8. Proof. We shall establish (18-8) by considering the case when A is acute, as in Fig. 18-8 (a) and the case when A is obtuse as in Fig. 18-8 (&). The case A = 90° involves no difficulty, and it will therefore be omitted. Let h denote the altitude from C to the side AB. Also, let x denote the length AD. Hence, DB is c — x in Fig. 18-8 (a) and is c + x in Fig. 18-8(6). Sec. 1 8-6 Solution of the General Triangle 297 In Fig. 18-8 (a), (c - x) 2 + h 2 = a 2 , and X 2 + h 2 = b 2 . Subtracting, we have c 2 — 2cx = a 2 — 6 2 , or a 2 = b 2 + c 2 - 2cz. a 2 = 6 2 + c 2 — 26c cos A. (c + a;) 2 + h 2 = a 2 , Since # = b cos A, In Fig. 18-8(6), and x 2 + h 2 = 6 2 . Subtracting, we find that c 2 + 2c# = a 2 — 6 2 , or a 2 = 6 2 + c 2 + 2cz. Since x = — 6 cos A in this case, a 2 = b 2 + c 2 — 26c cos A. By drawing altitudes to the other sides and proceeding in a similar manner, we obtain (18-9) and (18-10). Note. For a simple algebraic proof of the law of cosines, see prob- lem 22 in Exercise 18-4. The law of cosines applies equally well if ABC is a right triangle. In this case, one of the formulas reduces to the pythogorean theorem, since cos 90° = 0. 18-6. SOLUTION OF CASE III AND CASE IV BY THE LAW OF COSINES Since the law of cosines is expressed by formulas involving addi- tion and subtraction, it is not well adapted to logarithmic computa- tion and its use is not recommended unless the given sides are easily squared. Example 18-6. Given b = 9.0, c = 13.0, A = 115°10', solve the triangle ABC. Solution: By the law of cosines, a 2 = b 2 + c 2 - 26c cos A = 9 2 + (13) 2 - 2(9) (13) cos 115°10' = 81 + 169 - 234( - cos 64°50') = 250 + 234 (0.4253) = 349.52. Hence, a = 18.7. We employ the law of sines to find angle B. Thus, sin£ ^ sin 115°10' sin 64°50 / 9 18.7 18.7 and B = 25°50'. Therefore, C = 180° - (115°10' + 25°50') = 39°. 298 Solution of the General Triangle Sec. 1 8-6 Example 18-7. Given a = 3, 6 = 5, c = 7. Find the angles. Solution: From a 2 = 6 2 + c 2 — 26c cos ^4, we have the formula 6 2 + c 2 - a 2 COS A = -^ Therefore, _ ^ Hence, A = 21*47'. Similarly, B = 38°13' and C = 120°. We can check these by the equation A -f 5 -f C = 180°. EXERCISE 18-2 In each of the problems from 1 to 6, solve the given triangle by the law of cosines. 1. a = 300, 6 = 250, C - 58°40'. 2. a = 50, c = 240, 5 = 110°50'. 3. 6 = 65, c = 310, A = 67°10 / . 4. a = 130, c = 90, £ = 100°20'. 5. 6 = 50, c = 110, A = 150°. 6. a = 1.63, 6 = 3.45, C = 26°10'. 7. If a = 15, b = 12, c - 20, find A. 8. If a = 25, 6 = 30, and c = 35, find B. 9. If a = 100, 6 = 300, c = 500, find C. 10. If a = 15, 6 = 12, and c = 20, find B. 11. If a = 16, 6 = 17, and c = 18, find A, B y C. 12. If a = 260, 6 = 322, c = 481, find A, £, C. 13. The distance between two points A and B cannot be measured directly. Accordingly, a third point C is selected, and it is found that AC = 3000 feet, £C = 4500 feet, and angle ACB = 46°20'. Find the distance A£. 14. Two sides of a parallelogram are 125 feet and 200 feet, and the included angle is 110°30'. Find the length of the longer diagonal of the parallelogram and also the angle between that diagonal and a longer side of the parallelogram. 15. Two sides of a triangular plot of ground are 250 feet and 200 feet, and the included angle is 67°33'. Find the perimeter of the plot. 16. Two sides of a parallelogram are 700 feet and 420 feet, and one diagonal is 600 feet. Find the length of the other diagonal. 17. In a triangle ABC, a = 25, b = 27, and the median from A is 20. Find c, A, B, C. 18-7. THE LAW OF TANGENTS Law of Tangents. Let ABC be any triangle. Then the following relationships exist between two sides and the angles opposite them : . tan i (A - B) (18-n) ^z^ = 2 :, a + b tan - (A + B) Sec. 18-7 Solution of the General Triangle 299 (18-12) 6-c ta 4< B ~ C > (18-13) b + c tan ^ (B + C) tan J (C - A) c — a 2 ' C + a tan±(C + A) These relationships constitute the law of tangents. Proof. Let us denote the common ratio of the law of sines by r. Thus, a = r sin A, 6 = r sin J5, c = r sin C Then a — b __ r sin A — r sin B _ sin A — sin J? a + 6 ~~ r sin A + r sin B "" sin A + sin B Substituting from (7-27) and (7-28) of Section 7-4, we have smA+sinB 2 sin * (A + fi) cos 1 (4 _ £) Hence, . tan I (A - B) (18-11) ^l = — T^ "• a + 6 tani(A + B) A similar procedure may be followed to prove (18-12) and (18-13). We shall now use the law of tangents to solve a triangle in which two sides and the included angle are given. Note that this law is well adapted to logarithmic computation. Example 18-8. Given b = 249, c = 372, A = 56°22', solve the triangle ABC. Solution: To find C — B, we use the formula tan|(C-*)=^|tan|((7+B). Here c - b = 123, c + b = 621, C + B = 180° - A = 123°38', | (C + B) = 61°49'. Therefore, 19 „ tan±(C-£)=±gtan61°49', and - log tan i (C - B) = log 123 + log tan 61°49' - log 621. 300 Solution of f/ia Genera/ Triangle Sec. 1 8-7 The logarithmic work follows: log 123 = 2.0899 log tan 61°49' = 0.2710 12.3609 - 10 log 621 = 2.7931 log tan i (O - B) = 9.5678 - 10 Hence, and Check: i (C - B) = 20°17'. C=g(C+B)+|(C-B) = 82°6', £ = J (C 4- B) - ~ (C - B) = 41°32'. A +B +C = 56°22' + 82°6' + 41°32' = 180° To find a, we use the law of sines. Thus, 249sin56°22' 010 a== sin41°32' =313 - 18-8. THE HALF-ANGLE FORMULAS The following relationships are very convenient for the loga- rithmic solution of Case IV, where the three sides a, b, and c are known : (18-14) tan 2 = V s(s - a) » (18-15) tan 2 = y g(s - b) ' (18-16) ta C = ^( i -aH7^6) < 7 2 r s(s — c) In these formulas, 5 = - (a + b + c). Proof. From (7-16) in Section 7-3, we have , 9 A 1 — cos A tan J 75- = 7—: j • 2 1 + cos A Also, from the law of cosines, - b 2 + c 2 - a 2 COS it = ?TT • 26c Therefore ' , b 2 + c 2 - a 2 a 2 -(b-c) 2 (a + b - c) (a -b + c) 1 — cos A = 1 ^r = si = sr ' 26c 26c 26c and 1lMi _ , , 6 2 + c 2 - o 2 _ (6 + c) 2 - a 2 _ (b + c + a) (b + c - a) , l + oosil = l + ^ Wc Wc Sec. 18-8 Solution of the General Trhngle 301 If we let a + 6 + c = 2s, then a + b - c = a + b + c — 2c = 2(s — c), a - 6 + c = a + 6 + c - 26 = 2(s - 6), 6 + c — a = a + 6 + c — 2a = 2(s — a). Therefore, f o n 2 A - (<* + & - c) (a -- 6 + c) _ (s - c) (s - 6) ™ n 2 " (6 + c + a) (6 + c - a) " s(s - a) and (18-14) tanA = 4/IiZSZS. 2 r s(s — a) We can derive (18-15) and (18-16) in a similar manner. Example 18-9. Given: a = 379, b = 227, c = 416, find the angles of the triangle. Solution: Here 2s=a+b+c = 1022. Then 8 =511, s - a = 132, s - 6 = 284, s - c = 95. Hence, . A j / (284) (9 5) tan 2- = T ( 511)(132) ' The calculations by logarithms follow: log 284 = 2.4533 log 511 = 2.7084 log 95 = 1.9777 log 132 = 2.1206 log (284) (95) = 24.4310 - 20 4.8290 log (511) (132) = 4.8290 2 1 19.6020 - 20 log tan y = 9.8010 - 10 a . ., - A = 64°38'. Similarly, tanf = |/p»- tan 2 ~ V (511) (284) ' 2 ~ K ( 2 f (511) (95) Hence, we find that B = 32°46' and C = 82°36\ Check: A+B +C = 64°38' + 32°46' + 82°36' = 180°. EXERCISE 18-3 In each of the problems from 1 to 10, solve the given triangle by the law of tangents if an angle is given, or by the half-angle formulas if three sides are given 1. a = 50, 6 = 60, C = 60°. * 2. 6 = 17.1, c = 22.3, A = 21°16'. 3. o = 230, c = 106, B = 95°10'. 4. b = 79.3, c = 113, A = 133°14'1 302 Solution of the General Triangle Sec. 1 8-8 5. 6 = 41.82, c = 75.89, A = 78°49'. 6. a = 0.1028, 6 = 0.8726, C = 148°13\ 7. a = 625, 6 = 725, c = 825. 8. a = 60.65, 6 = 38.64, c = 23.57. 9. a = 67450, b = 84380, c = 98630. 10. a = 0.1146, 6 = 0.3184, c = 0.6379. 11. The diagonals of a parallelogram are 6 inches and 10 inches, and they intersect at an angle of 63°. Find the sides of the parallelogram. 12. Points A and B are separated by an obstacle. In order to find the distance between them, a third point C is selected and it is found that AC = 126 rods and BC = 185 rods. The angle subtended at C by AB is 96°14'. Find AB. 13. Two circles whose radii are 14 and 17 inches respectively intersect. The angle between the tangents to the circles at either point of intersection is 38°46'. Find the distance between the centers of the circles. 14. The sides of a parallelogram are 13.4 inches and 18.5 inches, and one diagonal is 15.6 inches. Find the angles and the other diagonal of the parallelogram. 15. Three circles whose radii are 10, 11, and 12 inches, respectively, are tangent to each other externally. Find the angles of the triangle formed by joining their centers. 16. The sides of a triangular field are in the proportion 4:5:6. The area of the field is 18 acres. If there are 160 square rods in an acre, find the length of each side of the field in rods. 17. In triangle ABC y prove the following: c 1 n cos-C a+b C0S l (A ~ B) c , 1 n sin -C These formulas are called Mollweide's equations. They may be used in checking the solution of a triangle. 18-9. AREA OF A TRIANGLE We can readily see that the area K of the triangle in Fig. 18-1 is (18-17) K = ^ch =^cb sin A. In either triangle, h = b sin A. In like manner, we obtain (18-18) K = lac sin B, (18-19) K = \ab sin C. 2 1 2° c sin B By substituting ^ for 6 from the law of sines, we may sin C transform (18-17) to obtain Sec. 1 8-9 Solution of the General Triangle 303 By cyclic interchanges of letters, we obtain (18-21) K = a2 sin B sinC (18-22) K = 2 sin A b 2 sin A sin C 2 sin B To derive a formula for finding the area of a triangle when its three sides are given, we first transform (18-17) in the following manner: K = ybc sin A = ^bc sin f 2 • -^-j 2 6c ( 2sin 2~ COS 2~) 2 , . A A = 6c sm y cos y But, from (7-14) and (7-15) in Section 7-3, sin A A /l — cos A , A A f\ - = j/ and cos y =4/- 2 Using the values from Section 18-8 for 1 — cos A and 1 + cos A, we have . A A /(s — c) (s — b) , A A /s(s — a) sm 2=V — Fc — - and C0S 2 = T be— Consequently, the formula for area in terms of the sides is (18-21) K = be |/ (S - C) 6 f ~ 6) V^HS = \/s(s — a) (s — b) (s — c). The following examples illustrate the use of the area formulas. Example 18-10. Find the area of the triangle in Example 18-8, in which b ='249, c = 372, A = 56°22'. Solution: Since two sides and the included angle are given, (18-17) may be used* Thus, K = he sin A = ^(249) (372) sin 56°22>. The logarithmic work follows : log 0.5 = 9.6990 - 10 log 249 = 2.3962 log 372 = 2.5705 log sin 56°22' = 9.9205 - 10 log K = 24.5862 - 20 K = 38,600. 304 Solution of the General Triangle Sec. 1 8-9 Example 18-11. Find the area of the triangle in Example 18-1, in which A = 38°14', B = 67°20', c = 329. Solution: Since two angles and a side are given, (18-20) may be used. In this case, „ c* sin A sin B _ (329)* sin 38°14' sin 67°20' _ QO inn * - 2 sinC ~ 2ih774 5 26 7 ~ 32 ' 100 ' Example 18-12. Find the area of the triangle in Example 18-9, in which a = 379, b = 227, c = 416. Solution: In this solution, (18-21) is used. We have K = V*(* -«)(*- &) (* - c) = V(5H) (132) (284) (95) = 42,700. EXERCISE 18-4 In each of the problems from 1 to 8, find the area of the given triangle. 1. a = 12.30, A = 36°25', B = 44°37'. 2. c = 461.3, B = 67°19 ; , C = 23°14'. 3. a = 987.4, 6 = 503.6, A = 54°13'. 4. 6 = 4.395, c = 9.806, C = 37°46'. 5. 6 = 65, c = 310, A = 67°10 ; . 6. a = 300, 6 = 250, C = 58°40'. 7. a = 15, 6 = 12, c = 20. 8. a = 100, 6 = 300, c = 500. 9. In triangle ABC, let r be the radius of the inscribed circle. Prove that K = rs and, therefore, that r = yi/ fr ~ a ) ( s ~ 6) ( s ""g) 10. Find the radius of the circle inscribed in the triangle whose sides are 48.92 feet, 63.86 feet, and 72.31 feet. 11. A cylindrical tank is to be built on a triangular lot having sides whose lengths are 200 feet, 186 feet, and 176 feet. Find the radius of the largest such tank which can be built on the lot. 12. In triangle ABC, let R be the radius of the circumscribed circle. Show that 2R= « = » = « . sin A sin B sin C 13. In triangle ABC, show that R = -rr? t where R is the radius of the circum- scribed circle and K is the area of the triangle. 14* The sides of a triangle are 23, 29, and 46 feet. Find the areas of the triangle and the inscribed and circumscribed circles. 15. The sides of a triangular plot of grass are 42 feet, 65 feet, and 87 feet. Find the minimum radius of action of an automatic lawn sprinkler which will water all parts of the plot from the same point. 16. An arc of a circle of radius r subtends a central angle 6. Show that the area bounded by this arc and its chord is jr r 2 (6 — sin 6). 17. Find the area of the largest pentagon which can be cut from a circular piece of metal 4 feet in radius. How much metal is wasted? 18. In triangle ABC, prove that the median from any vertex to the side opposite divides the angle at that vertex into two parts whose sines are proportional to the lengths of the parts into which the side opposite is divided by the median. Sec. 18-9 Solution of the General Triangle 305 19. In triangle ABC, prove that cos A , cos B , cos C a 2 + 6 2 + c 2 a b c 2abc 20. In triangle ABC, prove that a + b + c = (b + c) cos A + (c + a) cos B + (a + 6) cos C 21. In triangle ABC, prove that a 2 + b 2 + c 2 = a 2 (cos 2 C + sin 2 B) + 6 2 (cos 2 A + sin 2 C) + c 2 (cos 2 B + sin 2 A). 22. In triangle ABC, show that a = 6 cos C + c cos 5, 6 = c cos A + a cos C, c = a cos B -f 6 cos A. Multiply the first equation by a, the second by b, and the third by c, to give a second proof of the law of cosines; that is, prove (18-8) by showing that b 2 + c 2 - a 2 = 26c cos A. Similarly prove (18-9) and (18-10). 23. Consider any triangle ABC. If a > b, prove that A > B. If A > B, prove that a > b. 24. In triangles ABC and A'B'C, let A and A 1 , B and B', C and C be pairs of corresponding vertices, and let the corresponding sides be a and a', b and V, c and c'. If a = a', 6 = 6', and C > C", prove that c > c r . If a = a', 6 = &', and c > c', prove that C > C". Appendix A 30* TABLE I Four-Place Values of Functions op Numbers t sin t cos t tan t cot t sec t esc t .00 .0000 1.0000 .0000 1.000 .01 .0100 1.0000 .0100 99.997 1.000 100.66 .02 .0200 .9998 .0200 49.993 1.000 50.00 .03 .0300 .9996 .0300 33.323 1.000 33.34 .04 .0400 .9992 .0400 24.987 1.001 25.01 .05 .0500 .9988 .0500 19.983 1.001 20.01 .06 .0600 .9982 .0601 16.647 1.002 16.68 .07 .0699 .9976 .0701 14.262 1.002 14.30 .08 .0799 .9968 .0802 12.473 1.003 12.51 .09 .0899 .9960 .0902 11.081 1.004 11.13 .10 .0998 .9950 .1003 9.967 1.005 10.02 .11 .1098 .9940 .1104 9.054 1.006 9.109 .12 .1197 .9928 .1206 8.293 1.007 8.353 .13 .1296 .9916 .1307 7.649 1.009 7.714 .14 .1395 .9902 .1409 7.096 1.010 7.166 .15 .1494 .9888 .1511 6.617 1.011 6.692 .16 .1593 .9872 .1614 6.197 1.013 6.277 .17 .1692 .9856 .1717 5.826 1.015 5.911 .18 .1790 .9838 .1820 5.495 1.016 5.586 .19 .1889 .9820 .1923 5.200 1.018 5.295 .20 .1987 .9801 .2027 4.933 1.020 5.033 .21 .2085 .9780 .2131 4.692 1.022 4.797 .22 .2182 .9759 .2236 4.472 1.025 4.582 .23 .2280 .9737 .2341 4.271 1.027 4.386 .24 .2377 .9713 .2447 4.086 1.030 4.207 .25 .2474 .9689 .2553 3.916 1.032 4.042 .26 .2571 .9664 .2660 3.759 1.035 3.890 .27 .2667 .9638 .2768 3.613 1.038 3.749 .28 .2764 .9611 .2876 3.478 1.041 3.619 .29 .2860 .9582 .2984 3.351 1.044 3.497 .30 .2955 .9553 .3093 3.233 1.047 3.384 .31 .3051 .9523 .3203 3.122 1.050 3.278 .32 .3146 .9492 .3314 3.018 1.053 3.179 .33 .3240 .9460 .3425 2.920 1.057 3.086 .34 .3335 .9428 .3537 2.827 1.061 2.999 .35 .3429 .9394 .3650 2.740 1.065 2.916 .36 .3523 .9359 .3764 2.657 1.068 2.839 .37 .3616 .9323 .3879 2.578 1.073 2.765 .38 .3709 .9287 .3994 2.504 1.077 2.696 .39 .3802 .9249 .4111 2.433 1.081 2.630 t sin t cos t tan t cot t ' sec t CSC t i. 310 Appendix A TABLE I (continued) t sin t cos t tan t cot t sec t esc t .40 .3894 .9211 .4228 2.365 1.086 2.568 .41 .3986 .9171 .4346 2.301 1.090 2.509 .42 .4078 .9131 .4466 2.239 1.095 2.452 .43 .4169 .9090 .4586 2.180 1.100 2.399 .44 .4259 .9048 .4708 2.124 1.105 2.348 .45 .4350 .9004 .4831 2.070 1.111 2.299 .46 .4439 .8961 .4954 2.018 1.116 2.253 .47 .4529 .8916 .5080 1.969 1.122 2.208 .48 .4618 .8870 .5206 1.921 1.127 2.166 .49 .4706 .8823 .5334 1.875 1.133 2.125 .50 .4794 .8776 .5463 1.830 1.139 2.086 .51 .4882 .8727 .5594 1.788 1.146 2.048 .52 .4969 .8678 .5726 1.747 1.152 2.013 .53 .5055 .8628 .5859 1.707 1.159 1.978 .54 .5141 .8577 .5994 1.668 1.166 1.945 .55 .5227 .8525 .6131 1.631 1.173 1.913 .56 .5312 .8473 .6269 1.595 1.180 1.883 .57 .5396 .8419 .6410 1.560 1.188 1.853 .58 .5480 .8365 .6552 1.526 1.196 1.825 .59 .5564 .8309 .6696 1.494 1.203 1.797 .60 .5646 .8253 .6841 1.462 1.212 1.771 .61 .5729 .8196 .6989 1.431 1.220 1.746 .62 .5810 .8139 .7139 1.401 1.229 1.721 .63 .5891 .8080 .7291 1.372 1.238 1.697 .64 .5972 .8021 .7445 1.343 1.247 1.674 .65 .6052 .7961 .7602 1.315 1.256 1.652 .66 .6131 .7900 .7761 1.288 1.266 1.631 .67 .6210 .7838 .7923 1.262 1.276 1.610 .68 .6288 .7776 .8087 1.237 1.286 1.590 .69 .6365 .7712 .8253 1.212 1.297 1.571 .70 .6442 .7648 .8423 1.187 1.307 1.552 .71 .6518 .7584 .8595 1.163 1.319 1.534 .72 .6594 .7518 .8771 1.140 1.330 1.517 .73 .6669 .7452 .8949 1.117 1.342 1.500 .74 .6743 .7385 .9131 1.095 1.354 1.483 .75 .6816 .7317 .9316 1.073 1.367 1.467 .76 .6889 .7248 .9505 1.052 1.380 1.452 .77 .6961 .7179 .9697 1.031 1.393 1.437 .78 .7033 .7109 .9893 1.011 1.407 1.422 .79 .7104 .7038 1.009 .9908 1.421 1.408 t sint cos t tan t cot t sec t esc t Appendix A 311 TABLE I (continued) t sin t cos t tan t cot t sec t CSC t .80 .7174 .6967 1.030 .9712 1.435 1.394 .81 .7243 .6895 1.050 .9520 1.450 1.381 .82 .7311 .6822 1.072 .9331 1.466 1.368 .83 .7379 .6749 1.093 .9146 1.482 1.355 .84 .7446 .6675 1.116 .8964 1.498 1.343 .85 .7513 .6600 1.138 .8785 1.515 1.331 .86 .7578 .6524 1.162 .8609 1.533 1.320 .87 .7643 .6448 1.185 .8437 1.551 1.308 .88 .7707 .6372 1.210 .8267 1.569 1.297 .89 .7771 .6294 1.235 .8100 1.589 1.287 .90 .7833 .6216 1.260 .7936 1.609 1.277 .91 .7895 .6137 1.286 .7774 1.629 1.267 .92 .7956 .6058 1.313 .7615 1.651 1.257 .93 .8016 .5978 1.341 .7458 1.673 1.247 .94 .8076 .5898 1.369 .7303 1.696 1.238 .95 .8134 .5817 1.398 .7151 1.719 1.229 .96 .8192 .5735 1.428 .7001 1.744 1.221 .97 .8249 .5653 1.459 .6853 1.769 1.212 .98 .8305 .5570 1.491 .6707 1.795 1.204 .99 .8360 .5487 1.524 .6563 1.823 1.196 1.00 .8415 .5403 1.557 .6421 1.851 1.188 1.01 .8468 .5319 1.592 .6281 1.880 1.181 1.02 .8521 .5234 1.628 .6142 1.911 1.174 1.03 .8573 .5148 1.665 .6005 1.942 1.166 1.04 .8624 .5062 1.704 .5870 1.975 1.160 1.05 .8674 .4976 1.743 .5736 2.010 1.153 1.06 .8724 .4889 1.784 .5604 2.046 1.146 1.07 .8772 .4801 1.827 .5473 2.083 1.140 1.08 .8820 .4713 1.871 .5344 2.122 1.134 1.09 .8866 .4625 1.917 .5216 2.162 1.128 1.10 .8912 .4536 1.965 .5090 2.205 1.122 1.11 .8957 .4447 2.014 .4964 2.249 1.116 1.12 .9001 .4357 2.066 .4840 2.295 1.111 1.13 .9044 .4267 2.120 .4718 2.344 1.106 1.14 .9086 .4176 2.176 .4596 2.395 1.101 1.15 .9128 .4085 2.234 .4475 2.448 1.096 1.16 .9168 .3993 2.296 .4356 2.504 1.091 1.17 .9208 .3902 2.360 .4237 2.563 1.086 1.18 .9246 .3809 2.427 .4120 2.625 1.082 1.19 .9284 .3717 2.498 .4003 2.691 1.077 t sin t cos t tan t cot t sec t CSC t nt Appendix A TABLE I (continued) t sin t cos J tan t cot t sec t esc t 1.20 .9320 .3624 2.572 .3888 2.760 1.073 1.21 .9356 .3530 2.650 .3773 2.833 1.069 1.22 .9391 .3436 2.733 .3659 2.910 1.065 1.23 .9425 .3342 2.820 .3546 2.992 1.061 1.24 .9458 .3248 2.912 .3434 3.079 1.057 1.25 .9490 .3153 3.010 .3323 3.171 1.054 1.26 .9521 .3058 3.113 .3212 3.270 1.050 1.27 .9551 .2963 3.224 .3102 3.375 1.047 1.28 .9580 .2867 2.341 .2993 3.488 1.044 1.29 .9608 .2771 3.467 .2884 3.609 1.041 1.30 .9636 .2675 3.602 .2776 3.738 1.038 1.31 .9662 .2579 3.747 .2669 3.878 1.035 1.32 .9687 .2482 3.903 .2562 4.029 1.032 1.33 .9711 .2385 4.072 .2456 4.193 1.030 1.34 .9735 .2288 4.256 .2350 4.372 1.027 1.35 .9757 .2190 4.455 .2245 4.566 1.025 1.36 .9779 .2092 4.673 .2140 4.779 1.023 1.37 .9799 .1994 4.913 .2035 5.014 1.021 1.38 .9819 .1896 5.177 .1931 5.273 1.018 1.39 .9837 .1798 5.471 .1828 5.561 1.017 1.40 .9854 .1700 5.798 .1725 5.883 1.015 1.41 .9871 .1601 6.165 .1622 6.246 1.013 1.42 .9887 .1502 6.581 .1519 6.657 1.011 1.43 .9901 .1403 7.055 .1417 7.126 1.010 1.44 .9915 .1304 7.602 .1315 7.667 1.009 1.45 .9927 .1205 8.238 .1214 8.299 1.007 1.46 .9939 .1106 8.989 .1113 9.044 1.006 1.47 .9949 .1006 9.887 .1011 9.938 1.005 1.48 .9959 .0907 10.983 .0910 11.TJ29 1.004 1.49 .9967 .0807 12.350 .0810 12.390 1.003 1.50 .9975 .0707 14.101 .0709 14.137 1.003 1.51 .9982 .0608 16.428 .0609 16.458 1.002 1.52 .9987 .0508 19.670 .0508 19.695 1.001 1.53 .9992 .0408 24.498 .0408 24.519 1.001 1.54 .9995 .0308 32.461 .0308 32.476 1.000 1.55 .9998 .0208 48.078 .0208 48.089 1.000 1.56 .9999 .0108 92.620 .0108 92.626 1.000 1.57 1.0000 .0008 1255.8 .0008 1255.8 1.000 1.58 1.0000 -.0092 . -108.65 -.0092 -108.65 1.000 1.59 .9998 -.0192 -52.067 -.0192 -52.08 1.000 1.60 . .9996 -.0292 -34.233 -.0292 -34.25 1.000 t sin t cost tan t cot t sec t esc t Appendix A 313 TABLE II Four-Place Values of Functions Sin Cos Tan Cot Sec Gsc 0°00' 10' .0000 1.000 .0000 1.000 90° 00' 89° 50' 029 000 029 343.8 000 343.8 20' 058 000 058 171.9 000 171.9 40' 30' .0087 1.000 .0087 114.6 1.000 114.6 30' 40' 116 .9999 116 85.94 000 85.95 20' 0°50' lo 00' 10' 145 999 145 .0175 68.75 000 68.76 10' 89° 00* 88° 50' .0175 .9998 57.29 1.000 57.30 204 998 204 49.10 000 49.11 20' 233 997 233 42.96 000 42.98 40' 30' .0262 .9997 .0262 38.19 1.000 38.20 30' 40' 291 996 291 34.37 000 34.38 20' 1°50' 10' 320 995 320 31.24 001 31.26 10' 88° 00' 87° 50' .0349 .9994 993 .0349 378 28.64 1.001 28.65 378 26.43 001 26.45 20' 407 992 407 24.54 001 24.56 40' 30' .0436 .9990 .0437 22.90 1.001 22.93 30' 40' 465 989 466 21.47 001 21.49 20' 2° 50' 3° 00' 10' 494 988 495 20.21 001 20.23 10' 87° 00' 86° 50' .0523 .9986 .0524 19.08 18.07 1.001 19.11 552 985 553 002 18.10 20' 581 983 582 17.17 002 17.20 40' 30' .0610 .9981 .0612 16.35 1.002 16.38 30' 40' 640 980 641 15.60 002 15.64 20' 3° 50' 4° 00' 10' 669 978 670 14.92 002 14.96 10' 86° 00' 85*50' .0698 .9976 .0699 14.30 1.002 14.34 727 974 729 13.73 003 13.76 20' 756 971 758 13.20 003 13.23 40' 30' .0785 .9969 .0787 12.71 1.003 12.75 30' 40' 814 967 816 12.25 003 12.29 20' 4° 50' 50 00' 10' 843 964 846 11.83 004 11.87 10' 86° 00' 84° 50* .0872 .9962 .0875 11.43 1.004 11.47 901 959 904 11.06 004 11.10 20' 929 957 934 10.71 004 10.76 40' 30' .0958 .9954 .0963 10.39 1.005 10.43 30' 40' .0987 951 .0992 10.08 005 10.13 20? 5° 50' .1016 948 .1022 9.788 005 9.839 10' 84° W .1045 .9945 .1051 9.514 1.006 9.567 Cos Sin Cot Tan Csc Sec 314 Appendix A TABLE II (continued) Sin Cos Tan Cot Sec Csc e°oc ic .1045 .9945 .1051 9.514 1.006 9.567 84° 00' 83° 50' 074 942 080 255 006 309 2C 103 939 110 9.010 006 9.065 40' 30' .1132 .9936 .1139 8.777 1.006 8.834 30' 40 161 932 169 556 007 614 20' 6° 50' roc 10' 190 929 .9925 198 345 007 405 10' 83° 00' 82° 50' .1219 .1228 8.144 1.008 8.206 248 922 257 7.953 008 8.016 20' 276 918 287 770 008 7.834 40' 30' .1305 .9914 .1317 7.596 1.009 7.661 30' 40' 334 911 346 429 009 496 20' 7° 50' 8°0C 10' 363 907 376 269 009 337 10' 82° 00' 81° 50' .1392 .9903 .1405 7.115 1.010 7.185 7.040 421 899 435 6.968 010 20' 449 894 465 827 Oil 6.900 40' 30' .1478 .9890 .1495 6.691 1.011 6.765 30' 40' 507 886 524 561 012 636 20' 8° 50' 9° 00' 10' 536 881 554 435 012 512 10' 81° 00' 80° 50' .1564 .9877 872 .1584 6.314 1.012 6.392 593 614 197 013 277 20' 622 868 644 6.084 013 166 40' 30' .1650 .9863 .1673 5.976 1.014 6.059 30' 40' 679 858 703 871 014 5.955 20' 9° 50' 10° 00' 1C 708 853 733 769 015 855 10' 80° 00' 79° 50' .1736 765 .9848 .1763 5.671 1.015 5.759 843 793 576 016 665 20' 794 838 823 485 016 575 40' 30' .1822 .9833 .1853 5.396 1.017 5.487 30' 40' 851 827 883 309 018 403 20' 10° 50' 11° oc 10' 880 822 .9816 811 914 226 018 320 10' 79° 00' 78° 50' .1908 .1944 5.145 1.019 5.241 937 .1974 5.066 019 164 20' 965 805 .2004 4.989 020 089 40' 30' .1994 .9799 .2035 4.915 1.020 5.016 30' 40' .2022 793 065 843 021 4.945 20' 11° 50' 12° 00' 051 787 095 773 022 876 10' 78° 00' .2079 .9781 .2126 4.705 1.022 4.810 Cos Sin Cot Tan Csc Sec 4 ■ Appendix A 315 TABLE I] '. (continued) Sin Cos Tan Cot Sec Csc ia° oo' 10' .2079 .9781 .2126 4.705 1.022 4.810 78° 00' 77° 50' 108 775 156 638 023 745 20' 136 769 186 574 024 682 40' 30' .2164 .9763 .2217 4.511 1.024 4.620 30' 40' 193 757 247 449 025 560 20' 12° 50' 13° 00' 10' 221 750 278 390 026 502 10' 77° 00' 76° 50' .2250 .9744 .2309 4.331 1.026 4.445 278 737 339 275 027 390 20' 306 730 370 219 028 336 40' 30' .2334 .9724 .2401 4.165 1.028 4.284 30' 40' 363 717 432 113 029 232 20' 13° 50' 14° 00' 10' 391 710 462 .2493 061 030 182 10' 76° 00' 75° 50' .2419 .9703 4.011 1.031 4.134 447 696 524 3.962 031 086 20' 476 689 555 914 032 4.039 40' 30' .2504 .9681 .2586 3.867 1.033 3.994 30' 40' 532 674 617 821 034 950 20 7 14° 50' 15° 00' 10' 560 667 648 776 034 906 10' 75° 00' 74° 50' .2588 .9659 .2679 3.732 1.035 3.864 616 652 711 689 036 822 20' 644 644 742 647 037 782 40' 30' .2672 .9636 .2773 3.606 1.038 3.742 30' 40' 700 628 805 566 039 703 20' 15° 50' 16° 00' 10' 728 .2756 784 621 836 526 039 665 10' 74° 00' 73° 50' .9613 .2867 3.487 1.040 3.628 605 899 450 041 592 20' 812 596 931 412 042 556 40' 30' .2840 .9588 .2962 3.376 1.043 3.521 30' 40' 868 580 .2994 340 044 487 20' 16° 50' 17° 00' 10' 896 572 .3026 305 045 453 10' 73° 00' 72° 50' .2924 .9563 .3057 3.271 1.046 3.420 952 555 089 237 047 388 20' .2979 546 121 204 048 356 40' 30' .3007 .9537 .3153 3.172 1.049 3.326 30' 40' 035 528 185 140 049 295 20' 17° 50' 18° 00' 062 .3090 520 217 108 050 265 10' 72° W .9511 .3249 3.078 1.051 3.236 Cos Sin Cot Tan Csc Sec 316 Appendix A TABLE II (continued) Sin Cos Tan Cot Sec Csc 18° 00' 10' .3090 .9511 .3249 3.078 1.051 052 3.236 72*00' 71* 50' 118 502 281 047 207 20' 145 492 314 3.018 053 179 40' 30' .3173 .9483 .3346 2.989 1.054 3.152 30' 40' 201 474 378 960 056 124 20' 18*50' 19*00' 10' 228 465 411 932 057 098 10' 71* 00' 70* 50' .3256 .9455 .3443 2.904 1.058 3.072 283 446 476 877 059 046 20' 311 436 508 850 060 3.021 40' 30' .3338 .9426 .3541 2.824 1.061 2.996 30' 40' 365 417 574 798 062 971 20' 19* 50' 20*00' 10' 393 407 607 .3640 773 063 947 10' 70° 00' 69° 50' .3420 .9397 2.747 1.064 065 2.924 901 448 387 673 723 20' 475 377 706 699 066 878 40' 30' .3502 .9367 .3739 2.675 1.068 2.855 30' 40' 529 356 772 651 069 833 20' 20* 50' 21* 00' 10' 557 346 805 628 070 812 10' 69* 00' 68° 50' .3584 .9336 .3839 2.605 583 1.071 2.790 611 325 872 072 769 20' 638 315 906 560 074 749 40' 30' .3665 .9304 .3939 2.539 1.075 2.729 30' 40' 692 293 .3973 517 076 709 20' 21* 50' 22*00' 10' 719 283 .4006 496 077 689 10' 68* 00' 67° 50' .3746 .9272 .4040 2.475 1.079 2.669 773 261 074 455 080 650 20' 800 250 108 434 081 632 40' 30' .3827 .9239 .4142 2.414 1.082 2.613 30' 40' 854 228 176 394 084 595 20' 22* 50' 23* 00' 10' 881 216 210 375 085 577 10' 67* 00' 66* 50' .3907 .9205 .4245 2.356 1.086 2.559 934 194 279 337 088 542 20' 961 182 314 318 089 525 40' 30' .3987 .9171 .4348 2.300 1.090 2.508 30' 40' .4014 159 383 282 092 491 20' 23* 50' 24*00' 041 147 417 264 2.246 093 475 10' 66* 00' .4067 .9135 .4452 1.095 2.459 Cos Sin Cot Tan Csc Sec \ Appendix A 317 TABLE II (continued) Sin Cos Tan Cot Sec Csc 24" 00' 10' .4067 .9135 .4452 2.246 1.095 2.459 66" 00' 65" 50' 094 124 487 229 096 443 20' 120 112 522 211 097 427 40' 30' .4147 .9100 .4557 2.194 1.099 2.411 30' 40' 173 088 592 177 100 396 20' 24" 50' 25° 00' 10' 200 075 628 161 102 381 10' 65" 00' 64" 50' .4226 .9063 .4663 2.145 1.103 2.366. 253 051 699 128 105 352 20' 279 038 734 112 106 337 40' 30' .4305 .9026 .4770 2.097 1.108 2.323 30' 40' 331 013 806 081 109 309 20' 25" 50' 26" 00' 10' 358 .9001 841 066 111 295 10' 64" 00' 63" 50' .4384 .8988 975 .4877 2.050 1.113 2.281 410 913 035 114 268 20' 436 962 950 020 116 254 40' 30' .4462 .8949 .4986 2.006 1.117 2.241 30' 40' 488 936 .5022 1.991 119 228 20' 26° 50' 27° 00' 10' 514 923 059 .5095 977 121 215 10' 63" 00' 62" 50' .4540 .8910 1.963 949 1.122 2.203 566 897 132 124 190 20' 592 884 169 935 126 178 40' 30' .4617 .8870 .5206 1.921 1.127 2.166 30' 40' 643 857 243 907 129 154 20' 27" 50' 28" 00' 10' 669 843 280 894 131 142 10' 62" 00' 61" 50' .4695 .8829 .5317 1.881 1.133 2.130 720 816 354 868 134 118 20' 746 802 392 855 136 107 40' 30' .4772 .8788 .5430 1.842 1.138 2.096 30' 40' 797 774 467 829 140 085 20' 28" 50' 29" 00' 10' 823 760 505 816 142 074 10' 61" 00' 60" 50' .4848 .8746 .5543 581 1.804 1.143 2.063 874 732 792 145 052 20' 899 718 619 780 147 041 40' 30' .4924 .8704 .5658 1.767 1.149 2.031 30' 40' 950 689 696 756 151 020 20' 29" 50' 30" 00' .4975 675 735 744 153 010 10' 60" 00' .5000 .8660 .5774 1.732 1.155 2.000 Cos Sin Cot Tan Csc Sec 318 Appendix A TABLE II {continued) V Sin Cos Tan Cot Sec Csc 30° 00' 10' .5000 .8660 646 .5774 1.732 1.155 157 2.000 60° 00' 59° 50' 025 812 720 1.990 20' 050 631 851 709 159 980 40' 30' .5075 .8616 .5890 1.698 1.161 1.970 30' 40' 100 601 930 686 163 961 20' 30° 50' 31° 00' 10' 125 587 .8572 557 .5969 675 165 1.167 951 10' 59° 00' 58° 50' .5150 .6009 1.664 1.942 175 048 653 169 932 20' 200 542 088 643 171 923 40' 30' .5225 .8526 .6128 1.632 1.173 1.914 30' 40' 250 511 168 621 175 905 20' 31° 50' 32° 00' 10' 275 496 208 611 177 1.179 896 1.887 878 10' 58° 00' 57° 50' .5299 .8480 .6249 1.600 324 465 289 590 181 20' 348 450 330 580 184 870 40' 30' .5373 .8434 .6371 1.570 1.186 1.861 30' 40' 398 418 412 560 188 853 20' 32° 50' 33° 00' 10' 422 403 453 550 190 1.192 195 844 10' 57° 00' 56° 50' .5446 .8387 .6494 1.540 1.836 471 371 536 530 828 20' 495 355 577 520 197 820 40' 30' .5519 .8339 .6619 1.511 1.199 1.812 30' 40' 544 323 661 501 202 804 20' 33° 50' 34° 00' 10' 568 .5592 616 307 703 492 204 796 10' 66° 00' 55° 50' .8290 .6745 1.483 1.206 209 1.788 274 787 473 781 20' 640 258 830 464 211 773 40' 30 .5664 .8241 .6873 1.455 1.213 1.766 30' 40 688 225 916 446 216 758 20' 34° 50 35° 00' 10' 712 208 .6959 437 218 751 10' 55° 00' 54° 50' .5736 .8192 .7002 1.428 1.221 223 1.743 760 175 046 419 736 20' 783 158 089 411 226 729 40' 30' .5807 .8141 .7133 1.402 1.228 1.722 30' 40' 831 124 177 393 231 715 20' 35° 50' 36° 00' 854 107 221 385 233 708 10' 54° 00' .5878 Cos .8090 .7265 1.376 1.236 Cse 1.701 Sin Cot Tan Sec Appendix A 319 TABLE I] '. (continued) Sin .5878 Cos .8090 Tan .7265 Cot See Csc 36° 00' 10' 1.376 368 1.236 1.701 54° 00' 53° 50' 901 073 310 230 695 20' 925 056 355 360 241 688 40' 30' .5948 .8039 .7400 1.351 1.244 1.681 30' 40' 972 021 445 343 247 675 20' 36° 50' 37° 00' 10' .5995 .6018 .8004 490 335 249 668 10' 53° 00' 52° 50' .7986 .7536 1.327 1.252 1.662 041 969 581 319 255 655 20' 065 951 627 311 258 649 4C 30' .6088 .7934 .7673 1.303 1.260 1.643 30' 40' 111 916 720 295 263 636 20' 37° 50' 38° 00' 10' 134 898 766 288 266 630 10' 52° 00' 51° 50' .6157 .7880 .7813 1.280 1.269 1.624 180 862 860 272 272 618 20' 202 844 907 265 275 612 40' 30' .6225 .7826 .7954 1.257 1.278 1.606 30' 40' 248 808 .8002 250 281 601 20' 38° 50' 39° 00' 10' 271 790 050 242 284 595 10' 51° 00' 50° 50' .6293 .7771 .8098 1.235 1.287 1.589 316 753 146 228 290 583 20' 338 735 195 220 293 578 40' 30' .6361 .7716 .8243 1.213 1.296 1.572 30' 40' 383 698 292 206 299 567 20' 39° 50' 40° 00' 10 406 679 342 .8391 199 302 561 10' 50° 00 49° 50' .6428 .7660 1.192 1.305 1.556 450 642 441 185 309 550 20' 472 623 491 178 312 545 40' 30' .6494 .7604 .8541 1.171 1.315 1.540 30' 40' 517 585 591 164 318 535 20' 40° 50' 41° 00' 10' 539 566 .7547 528 642 157 322 529 10' 49° 00' 48° 50' .6561 .8693 1.150 1.325 1.524 583 744 144 328 519 20' 604 509 796 137 332 514 40' 30' .6626 .7490 .8847 1.130 1.335 1.509 30' 40' 648 470 899 124 339 504 20' 41° 50' 42° 00' 670 451 .8952 117 342 499 10' 48° 00* .6691 .7431 Sin .9004 Cot 1.111 1.346 1.494 Cos Tan Csc Sec 320 Appendix A TABLE II (continued) i Sin Cos Tan Cot Sec Csc 42 c >00' 10' .6691 .7431 .9004 1.111 1.346 1.494 48° OC 47° 50' 713 412 057 104 349 490 20' 734 392 110 098 353 485 40' 30' .6756 .7373 .9163 1.091 1.356 1.480 30' 40' 777 353 217 085 360 476 20' 42° 43° 50' 00' 10' 799 333 271 079 364 471 47° 00' 46° 50' .6820 .7314 .9325 1.072 1.367 1.466 841 294 380 066 371 462 20' 862 274 435 060 375 457 40' 30' .6884 .7254 .9490 1.054 1.379 1.453 30' 40' 905 234 545 048 382 448 20' 43° 44° 50' w 10' 926 214 601 042 386 444 10' 46° 00' 45° 50' .6947 .7193 .9657 1.036 1.390 1.440 967 173 713 030 394 435 20' .6988 153 770 024 398 431 40' 30* .7009 .7133 .9827 1.018 1.402 1.427 30' 40' 030 112 884 012 406 423 20' 44° 46° 50' 050 092 .7071 Sin .9942 006 410 418 10' 45° 00' .7071 Cos 1.000 1.000 1.414 Csc 1.414 Cot Tan See Appendix A 321 TABLE III Four-Place Logarithms of Numbers m 10 n 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 N 1 0043 0453 0828 1173 1492 1790 2068 2330 2577 2810 3032 3243 3444 3636 3820 3997 4166 4330 4487 4639 4786 4928 5065 5198 5328 5453 5575 5694 5809 5922 6031 1 8 0086 0492 0864 1206 1523 1818 2095 2355 2601 2833 3054 3263 3464 3655 3838 4014 4183 4346 4502 4654 4800 4942 5079 5211 5340 5465 5587 5705 5821 5933 6042 9 3 0128 0531 0899 1239 1553 1847 2122 2380 2625 2856 3075 3284 3483 3674 3856 4031 4200 4362 4518 4669 4814 4955 5092 5224 5353 5478 5599 5717 5832 5944 6053 3 4 0170 0569 0934 1271 1584 1875 2148 2405 2648 2878 3096 3304 3502 3692 3874 4048 4216 4378 4533 4683 4829 4969 5105 5237 5366 5490 5611 5729 5843 5955 6064 4 5 0212 0607 0969 1303 1614 1903 2175 2430 2672 2900 3118 3324 3522 3711 3892 4065 4232 4393 4548 4698 4843 4983 5119 5250 5378 5502 5623 5740 5865 5966 6076 6 0263 0646 1004 1335 1644 1931 2201 2455 2695 2923 3139 3345 3541 3729 3909 4082 4249 4409 4564 4713 4857 4997 5132 5263 5391 5514 5635 5752 5866 5977 6086 6 7 0294 0682 1038 1367 1673 1959 2227 2480 2718 2945 3160 3365 3560 3747 3927 4099 4266 4425 4579 4728 4871 5011 5145 5276 5403 5527 5647 5763 5877 5988 6096 7 8 0334 0719 1072 1399 1703 1987 2253 2504 2742 2967 3181 3386 3579 3766 3945 4116 4281 4440 4594 4742 4886 5024 5159 5289 6416 5539 5658 5775 5888 5999 6107 8 9 0374 0755 1106 1430 1732 2014 2279 2529 2765 2989 3201 3404 3598 3784 3962 4133 4298 4456 4609 4757 4900 5038 5172 5302 5428 5551 5670 5786 5899 6010 6117 9 .0000 .0414 .0792 .1139 .1461 .1761 .2041 .2304 .2553 .2788 .3010 .3222 .3424 .3617 .3802 .3979 .4150 .4314 .4472 .4624 .4771 .4914 .5051 .5186 .5315 .5441 .5563 .5682 .5798 .5911 .6021 5 322 Appendix A TABLE III (continued) V 40 41 42 43 44 48 46 47 48 49 80 51 52 53 54 68 56 57 58 59 60 61 62 63 64 68 66 67 68 69 70 N .6021 .6128 .6232 .6335 .6435 .6532 .6628 .6721 .6812 .6902 .6990 1 6031 6138 6243 6345 6444 6542 6637 6730 6821 6911 6998 7084 7168 7251 7332 7412 7490 7566 7642 7716 7789 7860 7931 8000 8069 8136 8202 8267 8331 8395 8457 1 2 6042 6149 6253 6355 6454 6551 6646 6739 6830 6920 7007 7093 7177 7259 7340 7419 7497 7574 7649 7723 7796 7868 7938 8007 8075 8142 8209 8274 8338 8401 8463 S 3 6053 6160 6263 6365 6464 6561 6656 6749 6839 6928 7016 7101 7185 7267 7348 7427 7506 7582 7657 7731 7803 7875 7945 8014 8082 8149 8215 8280 8344 8407 8470 8 4 6064 6170 6274 6375 6474 6571 6666 6758 6848 6937 7024 7110 7193 7275 7356 7435 7513 7589 7664 7738 7810 7882 7952 8021 8089 8156 8222 8287 8351 8414 8476 4 6 6075 6180 3284 6386 6484 6580 6675 6767 6857 6946 7033 7118 7202 7284 7364 7443 7520 7597 7672 7745 7818 7889 7959 8028 8096 8162 8228 8293 8357 8420 8482 6 6 6085 6191 6294 6395 6493 6590 6684 6776 6866 6955 7042 7126 7210 7292 7372 7451 7528 7604 7679 7752 7825 7896 7966 8035 8102 8169 8235 8299 8363 8426 8488 6 7 6096 6201 6304 6405 6603 6699 6693 6785 6875 6964 7050 7135 7218 7300 7380 7459 7536 7612 7686 7760 7832 7903 7973 8041 8109 8176 8241 8306 8370 8432 8494 7 8 6107 6212 6314 6415 6513 6609 6702 6794 6884 6972 7059 7143 7226 7308 7388 7466 7543 7619 7694 7767 7839 7910 7980 8048 8116 8182 8248 8312 8376 8439 8500 8 9 6117 6222 6325 6425 6522 6618 6712 6803 6893 6981 7067 7152 7235 7316 7396 7474 7551 7627 7701 7774 7846 7917 7987 8055 8122 8189 8254 8319 8382 8445 8506 9 .7076 .7160 .7243 .7324 .7404 .7482 .7559 .7634 .7709 .7782 .7853 .7924 .7993 .8062 .8129 .8195 .8261 .8325 .8388 .8451 Appendix A 323 TABLE III (continued) N 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 N .8451 1 8457 S 8463 3 8470 8531 8591 8651 8710 8768 8825 8882' 8938 8993 9047 9101 9154 9206 9258 9309 9360 9410 9460 9509 9557 9605 9652 9699 9745 9791 9836 9881 9926 9969 3 4 8476 5 8482 8543 8603 8663- 8722 8779 8837 8893 8949 9004 9058 9112 9165 9217 9269 9320 9370 9420 9469 9518 6 8488 8549 8609 8669 8727 8785 8842 8899 8954 9009 9063 9117 9170 9222 9274 9325 9375 9425 9474 9523 9571 9619 9666 9713 9759 9805 9850 9894 9939 9983 6 7 8494 8 8500 8561 8621 8681 8739 8797 8854 8910 8965 9020 9074 9128 9180 9232 9284 9335 9385 9435 9484 9533 9581 9 8506 8567 8627 8686 8745 8802 8859 8915 8971 9025 9079 9133 9186 9238 9289 9340 9390 9440 9489 9538 9586 9633 9680 9727 9773 9818 9863 9908 9952 9996 9 .8513 .8573 .8633 .8692 .8751 .8808 .8865 .8921 .8976 8519 8579 8639 8698 8756 8814 8871 8927 8982 9036 9090 9143 9196 9248 9299 9350 9400 9450 9499 9547 9595 9643 9689 9736 9782 9827 9872 9917 9961 1 8525 8585 8645 8704 8762 8820 8876 8932 8987 8537 8597 8657 8716 8774 8831 8887 8943 8998 8555 8615 8675 8733 8791 8848 8904 8960 9015 .9031 9042 9053 9069 .9085 .9138 .9191 .9243 .9294 .9345 .9395 .9445 .9494 9096 9149 9201 9253 9304 9355 9405 9455 9504 9106 9159 9212 9263 9315 9365 9415 9465 9513 9122 9175 9227 9279 9330 9380 9430 9479 9528 9576 9624 9671 9717 9763 9809 9854 9899 9943 9987 7 .9542 9552 9600 9647 9694 9741 9786 9832 9877 9921 9965 2 9562 9566 .9590 .9638 .9685 .9731 .9777 .9823 .9868 .9912 .9956 9609 9657 9703 9750 9795 9841 9886 9930 9974 4 9614 9661 9708 9754 9800 9845 9890 9934 9978 5 9628 9675 9722 9768 9814 9859 9903 9948 9991 8 324 Appendix A TABLE IV Four-Place Logarithms of Functions w L Sin L Tan LCot L Cos 10.0000 0°00' 90° 00' 10' 7.4637 7.4637 12.5363 .0000 89° 50' 20' .7648 .7648 .2352 .0000 40' 30' 7.9408 7.9409 12.0591 .0000 30' 40' 8.0658 8.0658 11.9342 .0000 20' 0°50' .1627 .1627 .8373 10.0000 10' 1°00' 8.2419 8.2419 11.7581 9.9999 89° 00' 10' .3088 .3089 .6911 .9999 88° 50' 20' .3668 .3669 .6331 .9999 40' 30' .4179 .4181 .5819 .9999 30' 40' .4637 .4638 .5362 .9998 20' 1° 50' .5050 .5053 .4947 .9998 10' a oo' 8.5428 8.5431 11.4569 9.9997 88° 00' 10' .5776 .5779 .4221 .9997 87° 50' 20' .6097 .6101 .3899 .9996 40' 30' .6397 .6401 .3599 .9996 30' 40' .6677 .6682 .3318 .9995 20' 2° 50' .6940 .6945 .3055 .9995 10' 3° 00' 8.7188 8.7194 11.2806 9.9994 87° 00' 10' .7423 .7429 .2571 .9993 86° 50' 20' .7645 .7652 .2348 .9993 40' 30' .7857 .7865 .2135 .9992 30' 40' .8059 .8067 .1933 .9991 20' 3° 50' .8251 .8261 .1739 .9990 10' 4° 00' 8.8436 8.8446 11.1554 9.9989 86° 00' 10' .8613 .8624 .1376 .9989 85° 50' 20' .8783 .8795 .1205 .9988 40' 30' .8946 .8960 .1040 .9987 30' 40' .9104 .9118 .0882 .9986 20' 4° 50' .9256 .9272 .0728 .9985 10' 5° 00' 8.9403 8.9420 11.0580 9.9983 85° 00' 10' .9545 .9563 .0437 .9982 84° 50' 20' .9682 .9701 .0299 .9981 40' 30' .9816 .9836 .0164 .9980 30' 40' 8.9945 8.9966 11.0034 .9979 20' 5° 50' 9.0070 9.0093 10.9907 .9977 10' 6° 00' 9.0192 9.0216 10.9784 9.9976 84° 00' L Cos L Cot L Tan LSin „^ „„ _,. Appendix A 325 TABLE IV (continued) . ^ L Sin 9.0192 L Tan 9.0216 .0336 L Cot 10.9784 .9664 L Cos 6° 00' 10' 9.9976 .9975 84° 00' 83° 50' .0311 20' .0426 .0453 .9547 .9973 40' 30' .0539 .0567 .9433 .9972 30' 40' .0648 .0678 .9322 .9971 20' 6° 50' .0755 .0786 .9214 .9969 10' 7° 00' 9.0859 9.0891 10.9109 9.9968 83° 00' 10' .0961 .0995 .9005 .9966 82° 50' 20' .1060 .1096 .8904 .9964 40' 30' .1157 .1194 .8806 .9963 30' 40' .1252 .1291 .8709 .9961 20' 7° 50' .1345 .1385 .8615 .9959 10' 8° 00' 9.1436 9.1478 10.8522 9.9958 82° 00' 10' .1525 .1569 .8431 .9956 81° 50' 20' .1612 .1658 .8342 .9954 40' 30' .1697 .1745 .8255 .9952 30' 40' .1781 .1831 .8169 .9950 20' 8° 50' .1863 .1915 .8085 .9948 10' 9° 00' 9.1943 9.1997 10.8003 9.9946 81° 00' 10' .2022 .2078 .7922 .9944 80° 50' 20' .2100 .2158 .7842 .9942 40' 30' .2176 .2236 .7764 .9940 30' 40' .2251 .2313 .7687 .9938 20' 9° 50' .2324 .2^89 .7611 .9936 10' 10° 00' 9.2397 9.2463 10.7537 9.9934 80° 00' 10' .2468 .2536 .7464 .9931 79° 50' 20' .2538 .2609 .7391 .9929 40' 30' .2606 .2680 .7320 .9927 30' 40' .2674 .2750 .7250 .9924 20' 10° 50' .2740 .2819 .7181 .9922 10' 11° 00' 9.2806 9.2887 10.7113 9.9919 79° 00' 10' .2870 .2953 .7047 .9917 78° 50' 20' .2934 .3020 .6980 .9914 40' 30' .2997 .3085 .6915 .9912 30' 40' .3058 .3149 .6851 .9909 20' 11° 50' .3119 .3212 .6788 .9907 10'. n°oo' 9.3179 9.3275 10.6725 9.9904 78° 00' L Cos L Cot L Tan L Sin 4 326 Appendix A TABLE IV (continued) V LSia 9.3179 L Tan 9.3275 LCot 10.6725 LCos 9.9904 12° 00' 78° 00' 10' .3238 .3336 .6664 .9901 77° 50' 20' .3296 .3397 .6603 .9899 40' 30' .3353 .3458 .6542 .9896 30' 40' .3410 .3517 .6483 .9893 20' 12° 50' .3466 .3576 .6424 .9890 10' 13° 00' 9.3521 9.3634 10.6366 9.9887 77° 00' 10' .3575 .3691 .6309 .9884 76° 50' 20' .3629 .3748 .6252 .9881 40' 30' .3682 .3804 .6196 .9878 30' 40' .3734 .3859 .6141 .9875 20' 13° 50' .3786 .3914 .6086 .9872 10' 14° 00' 9.3837 9.3968 10.6032 9.9869 76° 00' 10' .3887 .4021 .5979 .9866 75° 50' 20' .3937 .4074 .5926 .9863 40' 30' .3986 .4127 .5873 .9859 30' 40' .4035 .4178 .5822 .9856 20' 14° 50' .4083 .4230 .5770 .9853 10' 15° 00' 9.4130 9.4281 10.5719 9.9849 75° 00' 10' .4177 .4331 .5669 .9846 74° 50' 20' .4223 .4381 .5619 .9843 40' 30' .4269 .4430 .5570 .9839 30' 40' .4314 .4479 .5521 .9836 20' 15° 50' .4359 .4527 .5473 .9832 10' 16° 00' 9.4403 9.4575 10.5425 9.9828 74° 00' 10' .4447 .4622 .5378 .9825 73° 50' 20' .4491 .4669 .5331 .9821 40' 30' .4533 .4716 .5284 .9817 30' 40' .4576 .4762 .5238 .9814 20' 16° 50' .4618 .4808 .5192 .9810 10' 17° 00' 9.4659 9.4853 10.5147 9.9806 73° 00' 10' .4700 .4898 .5102 .9802 72° 50' 20' .4741 .4943 .5057 .9798 40' 30' .4781 .4987 .5013 .9794 30' 40' .4821 .5031 .4969 .9790 20' 17° 50' .4861 .5075 .4925 .9786 10' 18° 00' 9.4900 L Cos 9.5118 LCot 10.4882 LTan 9.9782 LSin 72° 00' Appendix A 327 TABLE IV (continued) , i... V. LSin I. Tan 9.5118 LCot 10.4882 L Cos 9.9782 I... pm 18° 00' 9.4900 72° 00' 10' .4939 .5161 .4839 .9778 71° 50' 20' .4977 .5203 .4797 .9774 40' 30' .5015 .5245 .4755 .9770 30' 40' .5052 .5287 .4713 .9765 20' 18° 50' .5090 .5329 .4671 .9761 10' 19° 00' 9.5126 9.5370 10.4630 9.9757 71° 00' 10' .5163 .5411 .4589 .9752 70° 50' 20' .5199 .5451 .4549 .9748 40' 30' .5235 .5491 .4509 .9743 30' 40' .5270 .5531 .4469 .9739 20' 19° 50' .5306 .5571 .4429 .9734 10' 20° 00' 9.5341 9.5611 10.4389 9.9730 70° 00' 10' .5375 .5650 .4350 .9725 69° 50' 20' .5409 .5689 .4311 .9721 40' 30' .5443 .5727 .4273 .9716 • 30' 40' .5477 .5766 .4234 .9711 20' 20° 50' .5510 .5804 .4196 .9706 10* 21° 00' 9.5543 9.5842 10.4158 9.9702 69° OC 10' .5576 .5879 .4121 .9697 68° sty 20' .5609 .5917 .4083 .9692 Aff 30' .5641 .5954 .4046 .9687 30* 40' .5673 .5991 .4009 .9682 20' 21° 50' .5704 .6028 .3972 .9677 10' 22° 00' 9.5736 9.6064 10.3936 9.9672 68° 00' 10' .5767 .6100 .3900 .9667 67° 50' 20' .5798 .6136 .3864 .9661 40' 30' .5828 .6172 .3828 .9656 30' 40' .5S59 .6208 .3792 .9651 20' 22° 50' .5889 .6243 .3757 .9646 10' 23° 00' 9.5919 9.6279 10.3721 9.9640 67° 00' 10' .5948 .6314 .3686 .9635 66° 50' 20' .5978 .6348 .3652 .9629 40' 30' .6007 .6383 .3617 .9624 30' 40' .6036 .6417 .3583 .9618 20' 23° 50' .6065 .6452 .3548 .9613 10' 24° 00' 9.6093 L Cos 9.6486 LCot 10.3514 LT*n 9.9607 LSin 66° 00' 328 Appendix A TABLE IV (continued) L Sin ( L Tan LCot 10.3514 L Cos 66*00' 24*00' 9.0093 9.6486 9.9607 10' .6121 .6520 .3480 .9602 65* 50' 20' .6149 .6553 .3447 .9596 40' 30' .6177 .6587 .3413 .9590 30' 40' .6205 .6620 .3380 .9584 20' 24° 50' .6232 .6654 .3346 .9579 10' 25° 00' 9.6259 9.6687 10.3313 9.9573 66*00' 10' .6286 .6720 .3280 .9567 64*50' 20' .6313 .6752 .3248 .9561 40' 30' .6340 .6785 .3215 .9555 30' 40' .6366 .6817 .3183 .9549 20' 25° 50' .6392 .6850 .3150 .9543 10' 2$° 00' 9.6418 9.6882 10.3118 9.9537 64° 00' 10' .6444 .6914 .3086 .9530 63* 50' 20' .6470 .6946 .3054 .9524 40' 30' .6495 .6977 .3023 .9518 30' 40' .6521 .7009 .2991 .9512 20' 26* 50' .6546 .7040 .2960 .9505 10' 27*00' 10' 9.6570 9.7072 .7103 10.2928 .2897 9.9499 .9492 63*00' 62* 50' .6595 20' .6620 .7134 .2866 .9486 40' 30' .6644 .7165 .2835 .9479 30' 40' .6668 .7196 .2804 .9473 20' 27* 50' .6692 .7226 .2774 .9466 10' 28*00' 9.6716 9.7257 10.2743 9.9459 62*00' 10' .6740 .7287 .2713 .9453 61* 50' 20' .6763 .7317 .2683 .9446 40' 30' .6787 .7348 .2652 .9439 30' 40' .6810 .7378 .2622 .9432 20' 28* 50' 29*00' 10' .6833 .7408 9.7438 .7467 .2592 10.2562 .2533 .9425 9.9418 .9411 10' 61*00' 60*50' 9.6856 .6878 20' .6901 .7497 .2503 .9404 40' 30' .6923 .7526 .2474 .9397 30* 40' .6946 .7556 .2444 .9390 2C 29*50' 30*00' .6968 .7585 9.7614 .2415 10.2386 LTan .9383 9.9375 L Sin lC 60*00' 9.6990 LCos LCot Appendix A TABLE IV {continued) 329 L Sin L Tan LCot L Cos 30° 00' 9.6990 9.7614 10.2386 9.9375 W ° 00' 10' .7012 .7644 .2356 .9368 59° 50' 20' .7033 .7673 .2327 .9361 40' 30' .7055 .7701 .2299 .9353 30' 40' .7076 .7730 .2270 .9346 20' 30° 50' .7097 .7759 .2241 .9338 10' 31° 00' 9.7118 9.7788 10.2212 9.9331 59° 00' 10' .7139 .7816 .2184 .9323 58° 50' 20' .7160 .7845 .2155 .9315 40' 30' .7181 .7873 .2127 .9308 30' 40' .7201 .7902 .2098 .9300 20' 31° 50' .7222 .7930 .2070 .9292 10' 32° 00' 9.7242 9.7958 10.2042 9.9284 58° 00' 10' .7262 .7986 .2014 .9276 57° 50' 20' .7282 .8014 .1986 .9268 40' 30' .7302 .8042 .1958 .9260 30' 40' .7322 .8070 .1930 .9252 20' 32° 50' .7342 .8097 .1903 .9244 10' 33° 00' 9.7361 9.8125 10.1875 9.9236 57° 00' 10' .7380 .8153 .1847 .9228 56° 50' 20' .7400 .8180 .1820 .9219 40' 30' .7419 .8208 .1792 .9211 30' 40' .7438 .8235 .1765 .9203 20' 33° 50' .7457 .8263 .1737 .9194 10' 34° 00' 9.7476 9.8290 10.1710 9.9186 56° 0C 10' .7494 .8317 .1683 .9177 55° 50' 20' .7513 .8344 .1656 .9169 40' 30' .7531 .8371 .1629 .9160 30' 40' .7550 .8398 .1602 .9151 20' 34° 50' .7568 .8425 .1575 .9142 10' 35° 00' 10' 9.7586 9.8452 .8479 10.1548 .1521 9.9134 .9125 55° 00' 54° 50' .7604 20' .7622 .8506 .1494 .9116 40' 30' .7640 .8533 .1467 .9107 30' 40' .7657 .8559 .1441 .9098 20' 35° 50' .7675 -8586 .1414 .9089 10' 36° 00' 9.7692 LCos 9.8613 LCot 10.1387 LTan 9.9080 64° 00' LSin 330 Appendix A TABLE IV (continued) , ^ LSin LTan LCot L Cos 36° 00' 10' 9.7692 .7710 9.8613 .8639 10.1387 .1361 9.9080 54° 00' 53° 50' .9070 20' .7727 .8666 .1334 .9061 40' 30' .7744 .8692 .1308 .9052 30' 40' .7761 .8718 .1282 .9042 20' 36° 50' 37° 00' .7778 9.7795 .8745 9.8771 .1255 10.1229 .9033 10' 53° 00' 9.9023 10' .7811 .8797 .1203 .9014 52° 50' 20' .7828 .8824 .1176 .9004 40' 30' .7844 .8850 .1150 .8995 30' 40' .7861 .8876 .1124 .8985 20' 37° 50' .7877 .8902 .1098 .8975 10' 38° 00' 9.7893 9.8928 10.1072 9.8965 52° 00' 10' .7910 .8954 .1046 .8955 51° 50' 20' .7926 .8980 .1020 .8945 40' 30' .7941 .9006 .0994 .8935 30' 40' .7957 .9032 .0968 .8925 20' 38° 50' .7973 .9058 .0942 .8915 10' 39° 00' 9.7989 9.9084 10.0916 9.8905 51° 00' 10' .8004 .9110 .0890 .8895 50° 50' 20' .8020 .9135 .0865 .8884 40' 30' .8035 .9161 .0839 .8874 30' 40' .8050 .9187 .0813 .8864 20' 39° 50' .8066 .9212 .0788 .8853 10' 40° 00' 9.8081 9.9238 10.0762 9.8843 50° 00' 10' .8096 .9264 .0736 .8832 49° 50' 20' .8111 .9289 .0711 .8821 40' 30' .8125 .9315 .0685 .8810 30' 40' .8140 .9341 .0659 .8800 20' 40° 50' .8155 .9366 .0634 .8789 10' 41° 00' 9.8169 9.9392 10.0608 9.8778 49° 00' 10' .8184 .9417 .0583 .8767 48° 50' 20' .8198 .9443 .0557 .8756 40' 30' .8213 .9468 .0532 .8745 30' 40' .8227 .9494 .0506 .8733 20' 41° 50' .8241 .9519 .0481 .8722 10' 42° 00' 9.8255 LCos 9.9544 10.0456 9.8711 48° 00' LCot L tan LSin ^ Appendix A 331 TABLE IV (continued) ^ LSin LTan L Cot L Cos 9.8711 42° 00' 9.8255 9.9544 10.0456 48° 00' 10' .8269 .9570 .0430 .8699 47° 50' 20' .8283 .9595 .0405 .8688 40' 30' .8297 .9621 .0379 .8676 30' 40' .8311 .9646 .0354 .8665 20' 42° 50' .8324 .9671 .0329 .8653 10' 43° 00' 9.8338 9.9697 10.0303 9.8641 47° 00' 10' .8351 .9722 .0278 .8629 46° 50' 20' .8365 .9747 .0253 .8618 40' 30' .8378 .9772 .0228 .8606 30' 40' .8391 .9798 .0202 .8594 20' 43° 50' .8405 .9823 .0177 .8582 10' 44° 00' 9.8418 9.9848 10.0152 9.8569 46° 00' 10' .8431 .9874 .0126 .8557 45° 50' 20' .8444 .9899 .0101 .8545 40' 30' .8457 .9924 .0076 .8532 30' 40' .8469 .9949 .0051 .8520 20' 44° 50' 45° 00' .8482 9.8495 L Cos .9975 .0025 10.0000 .8507 9.8495 10' 45° 00' 10.0000 LCot LTan LSin 332 Appendix A TABLE V Squares and Square Roots N N* Vn VlON N AP Vn VlON N N* Vn VlON 1.00 1.0000 1.00000 3.16228 1.60 2.5600 1.26491 4.00000 2.80 4.8400 1.48324 4.69042 1.01 1.0201 1.00499 3.17805 1.61 2.5921 1.26886 4.01248 2.21 4.8841 1.48661 4.70106 1.02 1.0404 1.00995 3.19374 1.62 2.6244 1.27279 4.02492 2.22 4.9284 1.48997 4.71169 1.03 1.0609 1.01489 3.20936 1.63 2.6569 1.27671 4.03733 2.23 4.9729 1.49332 4.72229 1.04 1.0816 1.01980 3.22490 1.64 2.6896 1.28062 4.04969 2.24 5.0176 1.49666 4.73286 1.05 1.1025 1.02470 3.24037 1.65 2.7225 1.28452 4.06202 2.25 5.0625 1.50000 4.74342 1.06 1.1236 1.02956 3.25576 1.66 2.7556 1.28841 4.07431 2.26 5.1076 1.50333 4.75395 1.07 1.1449 1.03441 3.27109 1.67 2.7889 1.29228 4.08656 2.27 5.1529 1.50665 4.76445 1.08 1.1664 1.03923 3.28634 1.68 2.8224 1.29615 4.09878 2.28 5.1984 1.50997 4.77493 1.09 1.1881 1.04403 3.30151 1.69 2.8561 1.30000 4.11096 2.29 5.2441 1.51327 4.78539 1.10 1.11 1.2100 1.04881 3.31662 1.70 1.71 2.8900 1.30384 4.12311 2.30 2.31 '5.2900 1.51658 4.79583 1.2321 1.05357 3.33167 2.9241 1.30767 4.13521 5.3361 1.51987 4.80625 1.12 1.2544 1.05830 3.34664 1.72 2.9584 1.31149 4.14729 2.32 6.3824 1.52316 4.81664 1.13 1.2769 1.06301 3.36155 1.73 2.9929 1.31529 4.15933 2.33 5.4289 1.52643 4.82701 1.14 1.2996 1.06771 3.37639 1.74 3.0276 1.31909 4.17133 2.34 5.4756 1.52971 4.83735 1.15 1.3225 1.07238 3.39116 1.75 3.0625 1.32288 4.18330 2.35 5.5225 1.53297 4.84768 1.16 1.3456 1.07703 3.40588 1.76 3.0976 1.32665 4.19524 2.36 5.5696 1.53623 4.85798 1.17 1.3689 1.08167 3.42053 1.77 3.1329 1.33041 4.20714 2.37 5 6169 1.53948 4.86826 1.18 1.3924 1.08628 3.43511 1.78 3.1684 1.33417 4.21900 2.38 5.6644 1.54272 4.87852 1.19 1.4161 1.09087 3.44964 1.79 3.2041 1.33791 4.23084 2.39 5.7121 1.54596 4.88876 1J0 1.21 1.4400 1.09545 3.46410 1.80 1.81 3.2400 1.34164 4.24264 2.40 2.41 5.7600 1.54919 4.89898 1.4641 1.10000 3.47851 3.2761 1.34536 4.25441 5.8081 1.55242 4.90918 1.22 1.4884 1.10454 3.49285 1.82 3.3124 1.34907 4.26615 2.42 5.8564 1.65663 4.91935 1.23 1.5129 1.10905 3.50714 1.83 3.3489 1.35277 4.27785 2.43 5.9049 1.55885 4.92950 1.24 1.5376 1.11355 3.52136 1.84 3.3856 1.35647 4.28952 2.44 5.9536 1.56205 4.93964 1.15 1.5625 1.11803 3.53553 1.85 3.4225 1.36015 4.30116 2.45 6.0025 1.56525 4.94976 1.26 1.6876 1.12250 3.54965 1.86 3.4596 1.36382 4.31277 2.46 6.0516 1.56844 4.95984 1.27 1.6129 1.12694 3.56371 1.87 3.4969 1.36748 4.32435 2.47 6.1009 1.57162 4.96991 1.28 1.6384 1.13137 3.57771 1.88 3.5344 1.37113 4.33590 2.48 6.1504 1.57480 4.97996 1.29 1.6641 1.13578 3.59166 1.89 3.6721 1.37477 4.34741 2.49 6.2001 1.57797 4.98999 1.80 1.31 1.6900 1.14018 3.60555 1.90 1.91 3.6100 1.37840 4.35890 2.50 2.51 6.2500 1.68114 6.00000 1.7161 1.14455 3.61939 3.6481 1.38203 4.37035 6.3001 1.58430 5.00999 1.32 1.7424 1.14891 3.63318 1.92 3.6864 1.38564 4.38178 2.52 6.3504 1.58745 6.01996 1.33 1.7689 1.15326 3.64692 1.93 3 7249 1.38924 4.39318 2.53 6.4009 1.59060 6.02991 1.34 1.7950 1.15758 3.66060 1.94 3.7036 1.39284 4.40454 2.54 6.4516 1.59374 5.03984 1.35 1.8225 1.16190 3.67423 1.95 3.8025 1.39642 4.41588 2.55 6.5025 1.59687 5.04975 1.36 1.8496 1.16619 3.68782 1.96 3.8416 1.40000 4.42719 2.56 6.5536 1.60000 5.05964 1.37 1.8769 1.17047 3.70135 1.97 3.8809 1.40357 4.43847 2.57 6.6049 1.60312 5.06952 1.38 1.9044 1.17473 3.71484 1.98 3.9204 1.40712 4.44972 2.58 6.6564 1.60624 5.07937 1.39 1.9321 1.17898 3.72827 1.99 3.9601 1.41067 4.46094 2.59 6.7081 1.60935 5.08920 1.40 1.41 1.9600 1.18322 3.74166 2.00 2.01 4.0000 1.41421 4.47214 2.60 2.61 6.7600 1.61245 5.09902 1.9881 1.18743 3.75500 4.0401 1.41774 4.48330 6.8121 1.61555 5.10882 1.42 2.0164 1.19164 3.76829 2.02 4.0804 1.42127 4.49444 2.62 6.8644 1.61864 6.11859 1.43 2.0449 1.19583 3.78153 2.03 4.1209 1.42478 4.50555 2.63 6.9169 1.62173 6.12835 1.44 2.0736 1.20000 3.79473 2.04 4.1616 1.42829 4.51664 2.64 6.9696 1.62481 5.13809 1.45 2.1025 1.20416 3.80789 8.06 4.2025 1.43178 4.52769 2.65 7.0225 1.62788 6.14782 1.46 2.1316 1.20830 3.82099 2.06 4.2436 1.43527 4.53872 2.66 7.0756 1.63095 5.15752 1.47 2.1609 1.21244 3.83406 2.07 4.2849 1.43875 4.54973 2.67 7.1289 1.63401 5.16720 1.48 2.1904 1.21655 3.84708 2.08 4.3264 1.44222 4.56070 2.68 7.1824 1.63707 5.17687 1.49 2.2201 1.22066 3.86005 2.09 4.3681 1.44568 4.57165 2.69 7.2361 1.64012 5.18652 1.60 1.51 2.2500 1.22474 3.87298 1.10 2.11 4.4100 1.44914 4.58258 2.70 2.71 7.2900 1.64317 6.19615 2.2801 1.22882 3.88587 4.4521 1.45258 4.59347 7.3441 1.64621 5.20577 1.52 2.3104 1.23288 3.89872 2.12 4.4944 1.45602 4.60435 2.72 7.3984 1.64924 5.21636 1.53 2.3409 1.23693 3.91152 2.13 4.5369 1.45945 4.61519 2.73 7.4529 1.66227 5.22494 1.54 2.3710 1.24097 3.92428 2.14 4.5796 1.46287 4.62601 2.74 7.5076 1.65529 5.23450 1.55 2.4025 1.24499 3.93700 8.15 4.6225 1.46629 4.63681 2.75 7.5625 1.65831 5.24404 1.56 2.4336 1.24900 3.94968 2.16 4.6656 1.46969 4.64758 2.76 7.6176 1.66132 5.25357 1.57 2.4649 1.25300 3.96232 2.17 4.7089 1.47309 4.65833 2.77 7.6729 1.66433 5.26308 1.58 2.4964 1.25698 3.97492 2.18 4.7524 1.47648 4.66905 2.78 7.7284 1.66733 5.27257 1.59 2.5281 1.26095 3.98748 2.19 4.7961 1.47986 4.67974 2.79 7.7841 1.67033 5.28205 1.50 N 2.5600 1.26491 4.00000 2.20 4.8400 1.48324 4.69042 2.80 7.8400 1.67332 5.29150 N* Vn Vvon N N* Vn VlON N - N» Vn VlON Appendix A 333 TA BLE V (cont inued) N N» Vn VlON N 8.40 N* Vn VlON 1 N 4.00 N* Vn Vmr 1.80 7.8400 1.67332 5.29150 11.5600 1.84391 5.83095 16.0000 2.00000 6.32456 2.81 7.8961 1.67631 5.30094 3.41 11.6281 1.84662 5.83952 4.01 16.0801 2.00250 6.33246 2.82 7.9524 1.67929 5.31037 3.42 11.6964 1.84932 5.84808 4.02 16.1604 2.00499 6.34035 2.83 8.0089 1.68226 5.31977 3.43 11.7649 1.85203 5.85662 4.03 16.2409 2.00749 6.34823 2.84 8.0656 1.68523 5.32917 3.44 11.8336 1.85472 5.86515 4.04 16.3216 2.00998 6.35610 1.65 8.1225 1.68819 5.33854 3.48 11.9025 1.85742 5.87367 44)6 16.4025 2.01246 6.36396 2.86 8.1796 1.69115 5.34790 3.46 11.9716 1.86011 5.88218 4.06 16.4836 2.01494 6.37181 2.87 8.2369 1.69411 5.35724 3.47 12.0409 1.86279 5.89067 4.07 16.5649 2.01742 6.37966 2.88 8.2944 1.69706 5.36656 3.48 12.1104 1.86548 5.89915 4.08 16.6464 2.01990 6.38749 2.89 8.3521 1.70000 5.37587 3.49 12.1801 1.86815 5.90762 4.09 16.7281 2.02237 6.39531 1.90 2.91 8.4100 170294 5.38516 3.60 3.51 12.2500 1.87083 5.91608 4.10 4.11 16.8100 2.02485 6.40312 8.4681 1.70587 5.39444 12.3201 1.87350 5.92453 6.9&06 16.8921 2.02731 6.41093 2.92 8.5264 1.70880 5.40370 3.52 12.3904 1.87617 4.12 16.9744 2.02978 6.41872 2.93 8.5849 1.71172 5.41295 3.53 12.4609 1.87883 5.94138 4.13 17.0569 2.03224 6.42651 2.94 8.6436 1.71464 5.42218 3.54 12.5316 1.88149 5.94979 4.14 17.1396 2.03470 6.43428 1.95 8.7025 1.71756 5.43139 3.55 12.6025 1.88414 5.95819 4.15 17.2225 2.03715 6.44205 2.96 8.7616 1.72047 5.44059 3.56 12.6736 1.88680 5.96657 4.16 17.3056 2.03961 6.44981 2.97 8.8209 1.72337 5.44977 3.57 12.7449 1.88944 5.97495 4.17 17.3889 2.04206 6.45755 2.98 8.8804 1.72627 5.45894 3.58 12.8164 1.89209 5.98331 4.18 17.4724 2.04450 6.46529 2.99 8.9401 1.72916 5.46809 3.59 12.8881 1.89473 5.99166 4.19 17.5561 2.04695 6.47302 3.00 3.01 9.0000 1.73205 5.47723 3.60 3.61 12.9600 1.89737 6.00000 4J0 4.21 17.6400 2.04939 6.48074 9.0601 1.73494 5.48635 13.0321 1.90000 6.00833 17.7241 2.05183 6.48845 3.02 9.1204 1.73781 5.49545 3.62 13.1044 1.90263 0.01664 4.22 17.8084 2.05426 6.49615 3.03 9.1809 1.74069 5.50454 3.63 13.1769 1.90526 6.02496 4.23 17.8929 2.05670 6.50384 3.04 9.2416 1.74356 5.51362 3.64 13.2496 1.90788 6.03324 4.24 17.9776 2.05913 6.51153 3.05 9.3025 1.74642 5.52268 3.66 13.3225 1.91050 6.04152 4J5 18.0625 2.06155 6.51920 3.06 9.3636 1.74929 5.53173 3.66 13.3956 1.91311 6.04979 4.26 18.1476 2.06398 6.52687 3.07 9.4249 1.75214 5.54076 3.67 13.4689 1.91572 6.05805 4.27 18.2329 2.06640 6.53452 3.08 9.4864 1.75499 5.54977 3.68 13.5424 1.91833 6.06630 4.28 18.3184 2.06882 6.54217 3.09 9.5481 1.75784 5.55878 3.69 13.6161 1.92094 6.07454 4.29 18.4041 2.07123 6.54981 3.10 3.11 9.6100 1.76068 5.56776 3.70 3.71 13.6900 1.92354 6.08276 4.80 4.31 18.4900 2.07364 6.55744 9.6721 1.70352 5.57674 13.7641 1.92614 6.09098 18.5761 2.07605 6.56506 3.12 9.7344 1.76635 5.58570 3.72 13.8384 1.92873 6.09918 4.32 18.6624 2.07846 6.57267 3.13 9.7969 1.76918 5.59464 3.73 13.9129 1.93132 6.10737 4.33 18.7489 2.08087 6.58027 3.14 9.8596 1.77200 5.60357 3.74 13.9876 1.93391 6.11555 4.34 18.8356 2.08327 6.58787 3.15 9.9225 1.77482 5.61249 8.75 14.0625 1.93649 6.12372 4.35 18.9225 2.08567 6.59545 3.16 9.9856 1.77764 5.62139 3.76 14.1376 1.93907 6.13188 4.36 19.0096 2.08806 6.60303 3.17 10.0489 1.78045 5.63028 3.77 14.2129 1.94165 6.14003 4.37 19.0969 2.09045 6.61060 3.18 10.1124 1.78326 5.63915 3.78 14.2884 1.94422 6.14817 4.38 19.1844 2.09284 6.61816 8.19 10.1761 1.78606 5.64801 3.79 14.3641 1.94679 6.15630 4.39 19.2721 2.09523 6.62671 3.10 3.21 10.2400 1.78885 5.65685 3.80 3.81 14.4400 1.94936 6.16441 4.40 4.41 19.3600 2.09762 6.63325 10.3041 1.79165 5.66569 14.5161 1.95192 6.17252 19.4481 2.10000 6.64078 3.22 10.3684 1.79444 5.67450 3.82 14.5924 1.95448 6.18061 4.42 19.5364 2.10238 6.64831 3.23 10.4329 1.79722 5.68331 3.83 14.6689 1.95704 6.18870 4.43 19.6249 2.10476 6.65582 3.24 10.4976 1.80000 5.69210 3.84 14.7456 1.95959 6.19677 4.44 19.7136 2.10713 6.66333 3.15 10.5625 1.80278 5.70088 3.85 14.8225 1.96214 6.20484 4.46 19.8025 2.10950 6.67083 3.26 10.6276 1.80555 5.70964 3.86 14.8996 1.96469 6.21289 4.46 19.8916 2.11187 6.67832 3.27 10.6929 1.80S31 5.71839 3.87 14.9769 1.96723 6.22093 4.47 19.9809 2.11424 6.68581 3.28 10.7584 1.81108 5.72713 3.88 15.0544 1.96977 6.22896 4.48 20.0704 2.11660 6.69328 3.29 10.8241 1.81384 5.73585 3.89 15.1321 1.97231 6.23699 4.49 20.1601 2.11896 6.70075 3.30 3.31 10.8900 1.81659 5.74456 3.90 3.91 15.2100 1.97484 6.24500 4.60 4.51 20.2500 2.12132 6.70820 10.9561 1.81934 5.75326 15.2881 1.97737 6.25300 20.3401 2.12368 6.71565 3.32 11.0224 1.82209 5.76194 3.92 15.3664 1.97990 6.26099 4.52 20.4304 2.12603 6.72309 3.33 11.0889 1.82483 5.77062 3.93 15.4449 1.98242 6.26897 4.53 20.5209 2.12838 6.73053 3.34 11.1556 1.82757 5.77927 3.94 15.5236 1.98494 6.27694 4.54 20.6116 2.13073 6.73795 8.35 11.2225 1.83030 5.78792 3.96 15.6025 1.98746 6.28490 4.55 20.7025 2.13307 6.74537 3.36 11.2896 1.83303 5.79655 3.96 15.6816 1.98997 6.29285 4.56 20.7936 2.13542 6.75278 3.37 113569 1.83576 5.80517 3.97 15.7609 1.99249 6.30079 4.57 20.8849 2.13776 6.76018 3.38 11.4244 1.83848 5.81378 3.98 15.8404 1.99499 6.30872 4.58 20.9764 2.14009 6.76757 3.39 11.4921 1.84120 5.82237 3.99 15.9201 1.99750 6.31664 4.59 21.0681 2.14243 6.77495 8.40 11.5600 1.84391 5.83095 4.00 N 16.0000 2.00000 6.32456 4.60 21.1600 2.14476 6.78233 N N* Vn Vxofi N* Vn Viw S iV» V* Vmr 334 Appendix A TABLE V {continued) N 4.60 N* Vn 2.14476 VlOAT N N* Vn 2.28035 Vion N N* Vn VlON 21.1600 6.78233 5 JO 27.0400 7.21110 5.80 33.6400 2.40832 7.61577 4.61 21.2521 2.14709 6.78970 5.21 27.1441 2.28254 7.21803 5.81 33.7561 2.41039 7.62234 4.62 21.3444 2.14942 6.79706 5.22 27.2484 2.28473 7.22496 5.82 33.8724 2.41247 7.62889 4.03 21.4369 2.15174 6.80441 5.23 27.3529 2.28692 7.23187 5.83 33.9889 2.41454 7.63544 4.64 21.5296 2:15407 6.81175 5.24 27.4576 2.28910 7.23878 5.84 34.1056 2.41661 7.64199 4.66 21.6225 2.15639 6.81909 5.25 27.5625 2.29129 7.24569 5.85 34.2225 2.41868 7.64853 4.66 21.7156 2.15870 6.82642 5.26 27.6676 2.29347 7.25259 5.86 34.3396 2.42074 7.65506 4.67 21.8089 2.16102 6.83374 5.27 27.7729 2.29565 7.25948 5.87 34.4569 2.42281 7.66159 4.68 21.9024 2.16333 6.84105 5.28 27.8784 2.29783 7.26636 5.88 34.5744 2.42487 7.66812 4.69 21.9961 2.16564 6.84836 5.29 27.9841 2.30000 7.27324 5.89 34.6921 2.42693 7.67463 4.70 4.71 22.0900 2.16795 6.85565 5.30 5.31 28.0900 2.30217 7.28011 5.90 5.91 34.8100 2.42899 7.68115 7.68765 22.1841 2.17025 6.86294 28.1961 2.30434 7.28697 34.9281 2.43105 4.72 22.2784 2.17256 6.87023 5.32 28.3024 2.30651 7.29383 5.92 35.0464 2.43311 7.69415 4.73 22.3729 2.17486 6.87750 5.33 28.4089 2.30868 7.30068 5.93 35.1649 2.43516 7.70065 4.74 22.4676 2.17715 6.88477 5.34 28.5156 2.31084 7.30753 5.94 35.2836 2.43721 7.70714 4.76 22.5625 2.17945 6.89202 5.35 28.6225 2.31301 7.31437 5.95 35.4025 2.43926 7.71362 4.76 22.6576 2.18174 6.89928 5.36 28.7296 2.31517 7.32120 5.96 35.5216 2.44131 7.72010 4.77 ?2.7529 2.18403 6.90652 5.37 28.8369 2.31733 7.32803 5.97 35.6409 2.44336 7.72658 4.78 22.8484 2.18632 6.91375 5.38 28.9444 2.31948 7.33485 5.98 35.7604 2.44540 7.73305 4.79 22.9441 2.18861 6.92098 5.39 29.0521 2.32164 7.34166 5.99 35.8801 2.44745 7.73951 4.80 4.81 23.0400 23.136? 2.19089 6.92820 5.40 5.41 29.1600 2.32379 7.34847 6.00 6.01 36.0000 2.44949 7.74597 2.19317 6.93542 29.2681 2.32594 7.35527 36.1201 2.45153 7.75242 4.82 23.2324 2.19545 6.94262 5.42 29.3764 2.32809 7.36206 6.02 36.2404 2.45357 7.75887 4.83 23.3289 2.19773 6.94982 5.43 29.4849 2.33024 7.36885 6.03 36.3609 2.45561 7.76531 4.84 23.4256 2.20000 6.95701 5.44 29.5936 2.33238 7.37564 6.04 36.4816 2.45764 7.77174 4.86 23.5225 2.20227 6.96419 5.45 29.7025 2.33452 7.38241 6.05 36.6025 2.45967 7.77817 4.86 23.6196 2.20454 6.97137 5.46 29.8116 2.33666 7.38918 6.06 36.7236 2.46171 7.78460 4.87 23.7169 2.20681 6.97854 5.47 29.9209 2.33880 7.39594 6.07 36.8449 2.46374 7.79102 4.88 23.8144 2.20907 6.98570 5.48 30.0304 2.34094 7.40270 6.08 36.9664 2.46577 7.79744 4.89 23.9121 2.21133 6.99285 5.49 30.1401 2.34307 7.40945 6.09 37.0881 2.46779 7.80385 4.90 4.91 24.0100 2.21359 7.00000 5.60 5.51 30.2500 2.34521 7.41620 6.10 6.11 37.2100 2.46982 7.81025 24.1081 2.21585 7.00714 30.3601 2.34734 7.42294 37.3321 2.47184 7.81665 4.92 24.2064 2.21811 7.01427 5.52 30.4704 2.34947 7.42967 6.12 37.4544 2.47386 7.82304 4.93 24.3049 2.22036 7.02140 5.53 30.5809 2.35160 7.43640 6.13 37.5769 2.47588 7.82943 4.94 24.4036 2.22261 7.02851 5.54 30.6916 2.35372 7.44312 6.14 37.6996 2.47790 7.83582 4.95 24.5025 2.22486 7.03562 5.55 30.8025 2.35584 7.44983 6.16 37.8225 2.47992 7.84219 4.96 24.6016 2.22711 7.04273 5.56 30.9136 2.35797 7.45654 6.16 37.9456 2.48193 7.84857 4.97 24.7009 2.22935 7.04982 5.57 31.0249 2.36008 7.46324 6.17 38.0689 2.48395 7.85493 4.98 24.8004 2.23159 7.05691 5.58 31.1364 2.36220 7.46994 6.18 38.1924 2.48596 7.86130 4.99 24.9001 2.23383 7.06399 5.59 31.2481 2.36432 7.47663 6.19 38.3161 2.48797 7.86766 6.00 5.01 25.0000 2.23607 7.07107 5.60 5.61 31.3600 2.36643 7.48331 6 JO 6.21 38.4400 2.48998 7.87401 25.1001 2.23830 7.07814 31.4721 2.36854 7.48999 38.5641 2.49199 7.88036 5.02 25.2004 2.24054 7.08520 5.62 31.5844 2.37065 7.49667 6.22 38.6884 2.49399 7.88670 5.03 25.3009 2.24277 7.09225 5.63 31.6969 2.37276 7.50333 6.23 38.8129 2.49600 7.89303 5.04 25.4016 2.24499 7.09930 5.64 31.8096 2.37487 7.50999 6.24 38.9376 2.49800 7.89937 5.06 25.5025 2.24722 7.10634 5.66 31.9225 2.37697 7.51665 6.25 39.0625 2.50000 7.90569 5.06 25.6036 2.24944 7.11337 5.66 32.0356 2.37908 7.52330 6.26 39.1876 2.50200 7.91202 5.07 25.7049 2.25167 7.12039 5.67 32.1489 2.38118 7.52994 6.27 39.3129 2.50400 7.91833 5.08 25.8064 2.25389 7.12741 5.68 32.2624 2.38328 7.53658 6.28 39.4384 2.50599 7.92465 5.09 25.9081 2.25610 7.13442 5.69 32.3761 2.38537 7.54321 6.29 39.5641 2.50799 7.93095 6.10 5.11 26.0100 2.25832 7.14143 6.70 5.71 32.4900 2.38747 7.54983 6.30 6.31 39.6900 2.50998 7.93725 26.1121 2.26053 7.14843 32.6041 2.38956 7.55645 39.8161 2.51197 7.94355 5.12 26.2144 2.26274 7.15542 5.72 32.7184 2.39165 7.56307 6.32 39.9424 2.51396 7.94984 5.13 26.3169 2.26495 7.16240 5.73 32.8329 2.39374 7.56968 6.33 40.0689 2.51595 7.95613 5.14 26.4196 2.26716 7.16938 5.74 32.9476 2.39583 7.57628 6.34 40.1956 2.51794 7.96241 5.16 26.5225 2.26936 7.17635 5.75 33.0625 2.39792 7.58288 6.35 40.3225 2.51992 7.96869 5.16 26.6256 2.27156 7.18331 5.76 33.1776 2.40000 7.58947 6.36 40.4496 2.52190 7.97496 5.17 26.7289 2.27376 7.19027 5.77 33.2929 2.40208 7.59605 6.37 40.5769 2.52389 7.98123 5.18 26.8324 2.27596 7.19722 $.78 33.4084 2.40416 7.60263 6.38 40.7044 2.52587 7.98749 5.19 26.9361 2.27816 7.20417 5.79 33.5241 2.40624 7.60920 6.39 40.8321 2.52784 7.99375 5.20 N 27.0400 2.28035 7.21110 5.60 33.6400 2.40832 7.61577 6.40 40.9600 2.52982 8.00000 AP Vn VTqn N iV» Vn VTqn N N* Vn vT<w Appendix A 335 TABLE V (cat itinued] ) N N* Vn VlON N 7.00 N* Vn VlON N 7.60 JV» Vn Vwf 6.40 40.9600 2.52982 8.00000 49.0000 2.64575 8.36660 57.7600 2.75681 8.71780 6.41 41.0881 2.53180 8.00625 7.01 49.1401 2.64764 8.37257 7.61 57.9121 2.75862 8.72353 6.42 41.2164 2.53377 8.01249 7.02 49.2804 2.64953 8.37854 7.62 58.0644 2.76043 8.72926 0.43 41.3449 2.53574 8.01873 7.03 49.4209 2.65141 8.38451 7.63 58.2169 2.76225 8.73499 6.44 41.4736 2.53772 8.02496 7.04 49.5616 2.65330 8.39047 7.64 58.3696 2.76405 8.74071 •.49 41.6025 2.53969 8.03119 7.06 49.7025 2.65518 8.39643 7.66 58.6225 2.76586 8.74643 6.46 41.7316 2.54165 8.03741 7.06 49.8436 2.65707 8.40238 7.66 58.6756 2.76767 8.75214 6.47 41.8609 2.54362 8.04363 7.07 49.9849 2.65895 8.40833 7.67 58.8289 2.76948 8.75785 6.48 41.9904 2.54558 8.04984 7.08 50.1264 2.66083 8.41427 7.68 58.9824 2.77128 8.76356 A.49 42.1201 2.54755 8.05605 7.09 50,2681 2.66271 8.42021 7.69 59.1361 2.77308 8.76926 6.50 6.51 42.2500 2.54951 8.06226 7.10 7.11 50.4100 2.66458 8.42615 7.70 7.71 59.2900 2.77489 8.77496 42.3801 2.55147 8.06846 50.5521 2.66646 8.4321)8 59.4441 2.77669 8.78066 6.52 42.5104 2.55343 8.07465 7.12 50.6944 2.66833 8.43801 7.72 59.6984 2.77849 8.78635 6.53 42.6409 2.55539 8.08084 7.13 50.8369 2.67021 8.44393 7.73 69.7529 2.78029 8.79204 6.54 42.7716 2.55734 8.06703 7.14 60.9796 2.67208 8.44985 7.74 59.9076 2.78209 8.79773 6.56 42.9025 2.55930 8.09321 7.15 51.1225 2.67395 8.45577 7.75 60.0625 2.78388 8.80341 6.56 43.0336 2.56125 8.09938 7.16 51.2656 2.67582 8.46168 7.76 60.2176 2.78568 8.80909 6.57 43.1649 2.56320 8.10555 7.17 51.4089 2.67769 8.46759 7.77 60.3729 2.78747 8.81476 6.58 43.2964 2.56515 8.11172 7.18 51.5524 2.67955 8.47349 7.78 60.5284 2.78927 8.82043 6.59 43.4281 2.56710 8.11788 7.19 51.6961 2.68142 8.47939 7.79 60.6841 2.79106 8.82610 6.60 6.61 43.5600 2.56905 8.12404 7 JO 7.21 51.8400 2.68328 8.48528 7.80 7.81 60.8400 2.79285 8.83176 43.6921 2.57099 8.13019 51.9841 2.68614 8.49117 60.9961 2.79464 8.83742 6.62 43.8244 2.57294 8.13634 7.22 52.1284 2.68701 8.49706 7.82 61.1524 2.79643 8.84308 6.63 43.9569 2.57488 8.14248 7.23 52.2729 2.68887 8.50294 7.83 61.3089 2.79821 8.84873 6.64 44.0896 2.57682 8.14862 7.24 52.4176 2.69072 8.50882 7.84 61.4656 2.80000 8.85438 6.M 44.22% 2.57876 8.15475 7J5 52.5625 2.69258 8.51469 7J6 61.6225 2.80179 8.86002 6.66 44.3556 2.58070 8.16088 7.26 52.7076 2.69444 8.52056 7.86 61.7796 2.80357 8.86566 8.87130 6.67 44.4889 2.58263 8.16701 7.27 52.8529 2.69629 8.62643 7.87 61.9369 2.80535 6.68 44.6224 2.58457 8.17313 7.28 52.9984 2.69815 8.53229 7.88 62.0944 2.80713 8.87694 6.69 44.7561 2.58650 8.17924 7.29 53.1441 2.70000 8.53815 7.89 62.2521 2.80891 8.88257 6.70 6.71 44.8900 2.58844 8.18535 7.30 7.31 53.2900 2.70185 8.54400 7.90 7.91 62.4100 2.81069 8.88819 45.0241 2.59037 8.19146 53.4361 2.70370 8.54985 62.5681 2.81247 8.89382 6.72 45.1584 2.59230 8.19756 7.32 53.5824 2.70555 8.55570 7.92 62.7264 2.81425 8.89944 6.73 45.2929 2.59422 8.20366 7.33 53.7289 2.70740 8.56154 7.93 62.8849 2.81603 8.90505 6.74 45.4276 2.59615 8.20975 7.34 53.8756 2.70924 8.56738 7.94 63.0436 2.81780 8.91067 ►6.70 45.5625 2.59808 8.21584 7.36 54.0225 2.71109 8.57321 7.96 63.2025 2.81957 8.91628 6.76 45.6976 2.60000 8.22192 7.36 54.1696 2.71293 8.57904 7.96 63.3616 2.82135 8.92188 6.77 45.8329 2.60192 8.22800 7.37 54.3169 2.71477 8.58487 7.97 63.5209 2.82312 8.92749 6.78 45.9684 2.60384 8.23408 7.38 54.4644 2.71662 8.59069 7.98 63.6804 2.82489 8.93308 6.79 46.1041 2.60576 8.24015 7.39 54.6121 2.71846 8.59651 7.99 63.8401 2.82666 8.93868 6.80 6.81 46.2400 2.60768 8.24621 7.40 7.41 54.760Q 2.72029 8.60233 8.00 8.01 64.0000 2.82843 8.94427 46.3761 2.60960 8.25227 54.9081 2.72213 8.60814 64.1601 2.83019 8.94986 6.82 46.5124 2.61151 8.25833 7.42 55.0564 2.72397 8.61394 8.02 64.3204 2.83196 2.83373 8.95545 6.83 46.6489 2.61343 8.26438 7.43 55.2049 2.72580 8.61974 8.03 64.4809 8.96103 6.84 46.7856 2.61534 8.27043 7.44 55.3536 2.72764 8.62554 8.04 64.6416 2.83549 8.96660 6.86 46.9225 2.61725 8.27647 7.46 55.5025 2.72947 8.63134 8.06 64.8025 2.83725 8.97218 6.86 47.0596 2.61916 8.28251 7.46 55.6516 2.73130 8.63713 8.06 64.9636 2.83901 8.97775 6.87 47.1969 2.62107 8.28855 7.47 55.8009 2.73313 8.64292 8.07 65.1249 2.84077 8.98332 6.88 47.3344 2.62208 8.29458 7.48 55.9504 2.73496 8.64870 8.08 65.2864 2.84253 8.98888 6.89 47.4721 2.62488 8.30060 7.49 56.1001 2.73679 8.65448 8.09 65.4481 2.84429 8.99444 6.90 6.91 47.6100 2.62679 8.30662 7.60 7.51 56.2500 2.73861 8.66025 8.10 8.11 65.6100 2.84605 9.00000 47.7481 2.62869 8.31264 56.4001 2.74044 8.66603 65.7721 2.84781 9.00555 6.92 47.8864 2.63059 8.31865 7.52 56.5504 2.74226 8.67179 8.12 65.9344 2.84956 9.01110 6.93 48.0249 2.63249 8.32466 7.53 56.7009 2.74408 8.67756 8.13 66.0969 2.85132 9.01665 6.94 48.1636 2.63439 8.33067 7.54 56.8516 2.74591 8.68332 8.14 66.2596 2.85307 9.02219 6.95 48.3025 2.63629 8.33667 7.66 67.0025 2.74773 8.68907 8.16 66.4225 2.85482 9.02774 6.96 48.4416 2.63818 8.34266 7.56 57.1536 2.74955 8.69483 8.16 66.5856 2.85657 9.03327 6.97 48.5809 2.64008 8.34865 7.57 57.3049 2.75136 8.70057 8.17 66.7489 2.85832 9.03881 6.98 48.7204 2.64197 8.35464 7.58 57.4564 2.75318 8.70632 8.18 66.9124 2.86007 9.04434 6.99 48.8601 2.64386 8.36062 7.59 57.6081 2.75500 8.71206 8.19 67.0761 2.86182 9.04986 7.00 N 49 0000 2.64575 8.36660 7.60 N 57.7600 2.75681 8.71780 8.20 AT 67.2400 2.86356 9.05539 N* VH Vudi N* Vn van N* Vn V$5R 336 Appendix A TABLE V (con tinued) H N* Vn viw? N JV« Vn vISat N N* Vii Vufir 8 JO 67.2400 2.86356 9.05539 8.80 77.4400 2.96648 9.38083 9.40 88.3600 3.06594 9.69536 8.21 67.4041 2.86531 9.06091 8.81 77.6161 2.96816 9.38616 9.41 88.5481 3.06757 9.70052 8.22 67.5684 2.86705 9.06642 8.82 77.7924 2.96985 9.39149 9.42 88.7364 3.06920 9.70567 8.23 67.7329 2.86880 9.07193 8.83 77.9689 2.97153 9.39681 9.43 88.9249 3.07083 9.71082 8.24 67.8976 2.87054 9.07744 8.84 78.1456 2.97321 9.40213 9.44 89.1136 3.07246 9.71597 8J5 68.0625 2.87228 908295 8.86 78.3225 2.97489 9.40744 9.45 89.3025 3.07409 9.72111 8.26 68.2276 2.87402 9.08845 8.86 78.4996 2.97658 9.41276 9.46 89.4916 3.07571 9.72625 8.27 68.3929 2.87576 9.09395 8.87 78.6769 2.97826 9.41807 9.47 89.6809 3.07734 9.73139 8.28 68.5584 2.87750 9.09945 8.88 78.8544 2.97993 9.42338 9.48 89.8704 3.07896 9.73653 8.29 68.7241 2.87924 9.10494 8.89 79.0321 2.98161 9.42868 9.49 90.0601 3.08058 9.74166 8.80 8.31 68.8900 2.88097 9.11043 8.90 8.91 79.2100 2.98329 9.43398 9.50 9.51 90.2500 3.08221 9.74679 69.0561 2.88271 9.11592 79.3881 2.98496 9.43928 90.4401 3.08383 9.75192 8.32 69.2224 2.88444 $.12140 8.92 79.5664 2.98664 9.44458 9.52 90.6304 3.08545 9.75705 8.33 69.3889 2.88617 9.12688 8.93 79.7449 2.98831 9.44987 9.53 90.8209 3.08707 9.76217 8.34 69.5556 2.88791 9.13236 8.94 79.9236 2.98998 9.45516 9.54 91.0116 3.08869 9.76729 8J6 69.7225 2.88964 9.13783 8.95 80.1025 2.99166 9.46044 9.65 91.2025 3.09031 9.77241 8.36 69.8896 2.89137 9.14330 8.96 80.2816 2.99333 9.46573 9.56 91.3936 3.09192 9.77753 8.37 70.0569 2.89310 9.14877 8.97 80.4609 2.99500 9.47101 9.67 91.5849 3.09354 9.78264 8.38 70.2244 2.89482 9.15423 8.98 80.6404 2.99666 9.47629 9.58 91.7764 3.09516 9.78775 8.39 70.3921 2.89655 9.15969 8.99 80.8201 2.99833 9.48156 9.59 91.9681 3.09677 9.79285 8.40 8.41 70.5600 2.89828 9.16515 9.00 9.01 81.0000 3.00000 9.48683 9.60 9.61 92.1600 3.09839 9.79796 70.7281 2.90000 9.17061 81.1801 3.00167 9.49210 92.3521 3.10000 9.80306 8.42 70.8964 2.90172 9.17606 9.02 81.3604 3.00333 9.49737 9.62 92.5444 3.10161 9.80816 8.43 71.0649 2.90345 9.18150 9.03 81.5409 3.00500 9.50263 9.63 92.7369 3.10322 9.81326 8.44 71.2336 2.90517 9.18695 9.04 81.7216 3.00666 9.50789 9.64 92.9296 3.10483 9.81835 8.45 71.4025 2.90689 9.19239 9.05 81.9025 3.00832 9.51315 9.65 93.1225 3.10644 9.82344 8.46 71.5716 2.90861 9.19783 9.06 82.0836 3.00998 9.51840 9.66 93.3156 3.10805 9.82853 8.47 71.7409 2.91033 9.20326 9.07 82.2649 3.01164 9.52365 9.67 93.5089 3.10966 9.83362 8.48 71.9104 2.91204 9.20869 9.08 82.4464 3.01330 9.52890 9.68 93.7024 3.11127 9.83870 8.49 72.0801 2.91376 9.21412 9.09 82.6281 3.01496 9.53415 9.69 93.8961 3.11288 9.84378 8.60 8.51 72.2500 2.91548 9.21954 9.10 9.11 82.8100 3.01662 9.53939 9.70 9.71 94.0900 3.11448 9.84886 72.4201 2.91719 9.22497 82.9921 3.01828 9.54463 94.2841 3.11609 9.85393 8.52 72.5904 2.91890 9.23033 9.12 83.1744 3.01993 9.54987 9.72 94.4784 3.11769 9.85901 8.53 72.7609 2.92062 9.23580 9.13 83.3569 3.02159 9.55510 9.73 94.6729 3.11929 9.86408 8.54 72.9316 2.92233 9.24121 9.14 83.5396 3.02324 9.56033 9.74 94.8676 3.12090 9.86914 8.65 73.1025 2.92404 9.24662 9.15 83.7225 3.02490 9.56556 9.76 95.0625 3.12250 9.87421 8.56 73.2736 2.92575 9.25203 9.16 83.9056 3.02655 9.57079 9.76 95.2576 3.12410 9.87927 8.57 73.4449 2.92746 9.25743 9.17 84.0889 3 02820 9.57601 9.77 95.4529 3.12570 9.88433 8.58 73.6164 2.92916 9.26283 9.18 84.2724 3.02985 9.58123 9.78 95.6484 3.12730 9.88939 8.59 73.7881 2.93087 9.26823 9.19 84.4561 3.03150 9.58645 9.79 95.8441 3.12890 9.89444 8.80 8.61 73.9600 2.93258 9.27362 9 JO 9.21 84.6400 3.03315 9.59166 9.80 9.81 96.0400 3.13050 9.89949 74.1321 2.93428 9.27901 84.8241 3.03480 9.59687 96.2361 3.13209 9.90454 8.62 74.3044 2.9359S 9.23440 9.22 85.0084 3.03645 9.60208 9.82 96.4324 3.13369 9.90959 8.63 74.4769 2.93769 9.28978 9.23 85.1929 3.03809 9.60729 9.83 96.6289 3.13528 9.91464 8.64 74.6496 2.93939 9.29516 9.24 85.3776 3.03974 9.61249 9.84 96.8256 3.13688 9.91968 8.68 74.8225 2.94109 9.30054 9J5 85.5625 3.04138 9.61769 9.85 97.0225 3.13847 9.92472 8.66 74.9956 2.94279 9.30591 9.26 85.7476 3.04302 9.62289 9.86 97.2196 3.14006 9.92975 8.67 75.1689 2.94449 9.31128 9.27 85.9329 3.04467 9.62808 9.87 97.4169 3.14166 9.93479 8.68 75.3424 2.94618 9.31665 9.28 861184 3.04631 9.63328 9.88 97.6144 3.14325 9.93982 8.69 75.5161 2.94788 9.32202 9.29 86.3041 3.04795 9.63846 9.89 97.8121 3.14484 9.94485 8.70 8.71 75.6900 2.94958 9.32738 9.80 9.31 86.4900 3.04959 9.64365 9.90 9.91 98.0100 3.14643 9.94987 75.8641 2.95127 9.33274 86.6761 3.05123 9.64883 98.2081 3.14802 9.95490 8.72 76.0384 2.95296 9.33809 9.32 86.8624 3.05287 9.65401 9.92 98.4064 3.14960 9.95992 8.73 76.2129 2.95466 9.34345 9.33 87.0489 3.05450 9.65919 9.93 98.6049 3.15119 9.96494 8.74 76.3876 2.95635 9.34880 9.34 87.2356 3.05614 9.66437 9.94 98.8036 3.15278 9.96995 8.75 76.5625 2.95804 9.35414 9.35 87.4225 3.05778 9.66954 9.95 99.0025 3.15436 9.97497 8.76 76.7376 2.95973 9.35949 9.36 87.6096 3.05941 9.67471 9.96 99.2016 3.15595 9.97998 8.77 76.9129 2.96142 9.36483 9.37 87.7969 3.06105 9.67988 9.97 99.4009 3.15753 9.98499 8.78 77.0884 2.96311 9.37017 9.38 87.9844 3.06268 9.68504 9.98 99.6004 3.15911 9.98999 8.79 77.2641 2.96479 9.37550 9.39 88.1721 3.06431 9.69020 9.99 99.8001 3.16070 9.99500 8.80 * 77.4400 2.96648 9.38083 9.40 N 88.3600 3.06594 9.69536 10.00 100.000 3.16228 10.0000 N* VN Vtoii N* VN VlON N N* VN Vioif EXERCISE 1-1. PAGE 11 1. (a) Commutative law of addition. (c) Associative law of multiplication. (e) Commutative law of multiplication. 2. (a) 2. (c) 1. (e) 1. (g) -5. (i) -2. (k) -7. 3. (a) -6. (c) -35. (e) -10. (g) 2. (i) 29. 4. (a) -5. (c) 0. (e) -2/3. (g) 36 - 2a. (i) -a6. (k) 3 - x. 5. (a)l. (c)5/17. (e) 1/1.02. <*> -™- <S)x+V- (k) r - 0.1. EXERCISE 1-2. PAGE 15 1. (a) -3, -2, 0, 4, 5. _(c) -4, -2, -1, -1/3, 10. (e) -2, -3/2, 1, y/3, 3. 2. (a) 3 > 1/3. (c) V2 > 1.414. (e) 22/7 > it. 3. (a) False, (c) True, (e) False. 4. (a) 0. (c) 0. (e) 2. (g) ^6. (i) 0. 5. (a) -1 < x < 1. (c) -a £ x ^ a. (e) -2 < x < 4. 6. (a) 4 < 5 < 6. (c) -1 < < 1. (e) 1 < y/Z < 2. 7. a + 7 ^ 10. EXERCISE 1-3. PAGE 20 1.32. 3.-1. 5.1,000,000. 7.-216. 9.2401. 11. —• 13. 16 ' 384 16 823,543 15.^- 17.32. 19.125a. 21. a\ 23. a 4 6 4 . 25. a 4 6°. 27. a". 29. a 2 »6 2 ™. 31. a 4 . 33. -• 35. —• 37. 5 - 6. 39. 46. 41. 3a - 46. a 4 a 4 43. -2a - 36. 45. (a - 2c)x 2 - (a + 26)*?/ + (6 - l)y 2 . 47. 4a - 56 - 2c. 49. a + (6 + c), a - (-6 - c). 51. a 2 + (c 2 - 6 2 ), a 2 - (6 2 - c 2 ). 53. 2a + (6 - 3c), 2a - (3c - 6). 55. x 2 + (- y 2 - z 2 ), x 2 - (y 2 + z 2 ). 57. -a 3 6 + (a& + 6 2 ), -a 3 & - (- a6 - 6 2 ). 59. 2.r + (- Sy - 4*), 2x - (3?/ + 4s). 61. 12. 63. 20. 65. 17. 67. 1. 69. -6. 71. 8. 73. 10. 75. 13. 77. 13. 79. |. 81. —• 7 14 EXERCISE 1-4. PAGE 24 I. 2.n/. 3. x 2 y 2 . 5. 5a 2 - 66 2 . 7. -4a. 9. a: 2 + 2xy. II. z 3 + Sx 2 - 3.r + 2. 13. -4xy. 15. 5a: 2 // 2 . 17. 3a 2 + 26 2 . 19. -2a + 26 - 2c. 21. x 2 - 2xy + y 2 . 23. ^ a: 3 - ^ s 2 + ~ xy + y y*. 25. -18a;?/. 27. -lOa; 3 ?/ 2 . 29. 24a 4 /> 2 c. 31. -6a; 2 - 4xy. 33. 12x 2 y 2 + Gxy*. 35. 2a; 2 + xy - 3?y 2 . 37. 4.? 4 - 8x*y - 4.r 2 ?/ 2 . 39. a; 3 - 2.r?/ 2 + 1/ 3 . 41. -2x. 43. 2x*/. 45. -1. 47. 2. 49. -zy + 2x - 3?/. 51. x + 5. 53. * + y. 55. s 8 + x 2 y + xy 2 + */ 3 . 57. x 2 - xy + y 2 . 59. .r 2 + .ry + y 2 . 61. 2y - xy. 63. x 3 + 3x 2 + Sx + 1 ; x 2 + 2x + 1. 65. 2.r 4 - 9a; + 6. 67. Odd values. 339 340 Appendix 8 EXERCISE 1-5. PAGE 30 1. 3(x + 2y). 3. 2(2x + 7). 5. a(3x + 2). 7. -x(a - 2c + x). 9. a(x - 2y + 3*). 11. ?/ 2 (5 + 3?/ - a). 13. (x - 4) (x + 4). 15. (2* - 3) (2x + 3). 17. (7x - 11) (7x -f- 11). 19. (3?/ - a) (3?/ + a). 21. a 2 x 2 (x - 3a) (x + 3a). 23. (0.1 - 6) (0.1 + b). 25. (7x*/ - 12a6) (7xy + 12a&). 27. x(6 + x 2 ) (6 - x 2 ). 29. 2(x + 2) (x 2 - 2x + 4). 31. (xy + z 2 ) (x 2 y 2 - xyz 2 + * 4 ). 33. (5p 2 a 3 + r 6 ) (25p 4 g 6 - 5p 2 ry 3 r 5 + r 10 ). 35. (x + y 2 ) (x - ?/ 2 ) (x 4 + x 2 y* + ?/ 8 ). 37. 3[3x 2 - (2y + 3z)] [9x 4 + 3x 2 (2?/ + 3e) + (2y + 3z) 2 ]. 39. (x + y - 2i + w) [(x + ?y) 2 + (x -f y) (z - iv) + (z - w) 2 ]. 41. (6x - y) (36x 2 + Qxy + 2/ 2 ). 43. x(6 - x 2 ) 2 . 45. (xy - 9) 2 . 47. (x 2 - 5) (x 2 - 3). 49. (4x - 3) (2x + 1). 51. (x - 2y) 2 . 53. (x - 4) (x + 3). 55. (x 2 + 1) (x 2 + 2). 57. (1 + 15x 3 ) 2 . 59. 5x(x + 4) (x - 2). 61. (2x + 1) (x - 6). 63. (2x - 1) (x + 3\ 65. (6x - 1) (x - 6). 67. (x - 0.6) 2 . 69. (a + 3) f.r + 2y). 71. (4x 2 - 5) (2x - 3). 73. (x + 3) (2a + y - z). 75. (x 2 - 3) (x + 1). EXERCISE 1-6. PAGE 33 1. 2. 3. 1. 5. x - y. 1. 3x 3 ?/. 9. 2xy*z. 11. 1. 13. 24. 15. 24xyz. 17. 36x«y*z*. 19. x 2 - 4. 21. (x 2 - 49) (x - 3) (x 3 - S) (x 2 - 4). EXERCISE 1-7. PAGE 35 1. (a) 12. (c) 4a 2 . (e) 18x// 3 . (g) x - 1. (i) (a - x) (h + x). n , v 101 , v 3x 2 y , v (>x?y 3 , . 1 ,. x 3x - 7 ,, . 3x + 2 , x x 2 - (y - 3) 2 , x , x 1 , N 3a + 6 , x x Wiq^' «»*-Jfo. <«M- < s > 2^zv (u) i+x-2/ EXERCISE 1-8. PAGE 38 - 17 94 „ 57 _ -2 n x 3 - 3x 2 - 6 lt x 2 + 3x - 5 L 24 # 3 -~2T 5 -~20' 7 -xT3T' 9 ' x 3 = 1 1L ii 1Q x 2 + x + 4 3x 2 + 12x + 5 1 + x 2 (x + 1) (x - l) 2 ' (x + 2) (x + 3) (x + 4) ' 1 - x " 10 - x(x + by) 2(a 2 - ab ± b 2 ) 2x + y±3 9 _ x 2 -xy+ y 2 19. t ■ r-7 rr • 21. ; j~z • 16* • J5. r r « (x + y) (x - i/) 2 a 2 - b 2 xy x 2 - y 2 EXERCISE 1-9. PAGE 41 54 x 6 a Q 7a 3 6 4 3x ? /(x - 1) (x - 1) (x + 2) (9x 2 + 6x + 4) (2x - 5) (4x + !)(«- 8) 1A (x - 7) (x - 2) ' 1& - 9x 2 - 4 ' 1/# (5x + 1) (x - 5) ' Appendix 8 341 -3(4* +5) (2s +3), 28. 70. 3^. 42/M4* 2 + 5) 5 59 5z ».^ + *. +rt . » ";; T ;_ + T M> - «. l -^£&- 43. y f + i 2a \ - 45. (*+?)*. 2/(2/ + &) EXERCISE 1-10. PAGE 47 1. a; = -5/2. 3. x = -3/7. 5. a; = -2. 7. -rr 2 - 9.-11. 11.7. 13.10. lo 15.-2. 17.?/= — ^ 19. 2/= -y-" 21. y=— g— • _ 9r 4- 17 —71 —7 23.?/= , ■ 25.-5. 27.7. 29.1/4. 31.3. 33.-^. 35.-/. 4 2 4 9*37 4-^8 37. =£■• 39. ^. 41. 4. 43. ~- 45. 2. 47. £-. 49. 40, 58. 51. 6, 7. 46 11 28 41 ' 53. 921,600 sq ft. 55. 170 adults, 330 children. 57. 20°, 40°, 120°. EXERCISE 2-1. PAGE 52 2. (a) 5. (c) 4. (e) V34. (g) V§9. 3. (a) 5. (c) V2. (e) 7. (g) 4. (i) v^T&l (k) 2. EXERCISE 2-2. PAGE 56 I. The area of a circle is a function of the radius of the circle, A = 7rr 2 . 3. The area of a trapezoid is a function of its altitude and bases, A = ~ (&i + W. Z 5. The volume of a cylinder is a function of its height and the radius of its base, V = wr 2 h. 7. The annual premium of a life insurance policy is a function of the applicant's age and physical condition, of the type of policy, of the company's rate policy, etc. No formula can be written. 9. -3, -5, 3, -2, 2 V2 -3, -3, 3/4, 2y - 3, \ - 3, ^—g- II. - ~ , 288, 6?/ 2 + 7. 13. 0. 15. 14. 17. v 2 + q 2 + r 2 . 19. 0. 21. 30. 23. 5/7. 25. 0. 27. A = nr', C = 2xr, A = ^- , C = 2 \ArX 29. £ = ^/36tt F 2 . 31. all .r. 33. all a\ 35. all x. 37. all a;. 39. all x. 41. all x. 43. all a?. 45. x * 0, 1. 47. | a; | £ 3. 49. a: = 0. 51. all a;. 53. x * 0, -2. 55. all a\ 57. a: ^ -1. 59. -2^xg2. EXERCISE 2-3. PAGE 57 1. 3,2.3,1*1. 3. 0,0.5. 5. 1,0,1,0,1. 342 Appendix 6 EXERCISE 2-4. PAGE 59 l.lf=yi. 3. 567. 5. 5. 7. 10/3. 9. -2. 13. 9/4, 27/8. 15. 1000/1, 100/1. 17. 327°C. 19. 99.5 lb, 95.2 lb. 21. 0.0324 in. EXERCISE 3-1. PAGE 67 1. (a) (1, 0). (c) (0, 1). (e) (1, 0). 2. (a) (0.54, 0.84). (c) (-0.99, 0.14). (e) (-0.65, -0.7G). 3. (a) V5/2. (c) 1. (e) -1/2. (g) -2. (i) - V3/2. 7. sin t COS t tan t cot t sec t esc £. (a) - ± V3/2 ± 1/V3 ±VS ± 2/y/S 2. (c) ±5/13 12/13 =b 5/12 db 12/5 — =fc 13/5. (e) ± V3/2 -1/2 db V§ ± 1/V3 — ±2A/3. (g) ±l/v<5 ±2/v / 5 1/2 — ± V5/2 ± Vs. (0 - ±4/5 ± 3/4 ± 4/3 ± 5/4 EXERCISE 3-3. PAGE 74 -5/3. 1. (a) 1. (c) 0.58. (e) 0.86. 2. (a) 0.9927. (c) 24.52. (e) -1.500. (g) 0.2571. (i) 5.798. (k) 1.011. EXERCISE 3-4. PAGE 80 1. tt/3. 3. tt/6. 5. 2tt/3. 7. tt/15. 9. 4tt/3. 11. 2tt/5. 13. 43tt/36. 15. 107tt/60. 17. 7tt/20. 19. 0.8090. 21. 1.4358. 23. 3.2107. 25. 1.6323. 27. 45°. 29. 270°. 31. 15°. 33. 630°. 35. 27°. 37. 470°23'54". 39. 43°43'. 57. 14tt/3, 1.25 radians. 59. 14.74 in. 61. (a) 4 radians, (b) 16/9 radians, (c) 0.04 radian. EXERCISE 3-5. PAGE 85 1. 0.5925. 3. 1.092. 5. 0.7412. 7. -1.453. 9. -0.2462. 11. 0.2504. 13. 1.181. 15. 9.010. 17. 0.6817. 19. 0.8437. 21. 0.9831. 23. 0.5154. 25.0.2930. 27.-9.462. 29.-1.059. 31. 173°10', 353°10'. 33. 42°30', 222°30'. 35. 3°10', 183°10'. 37. 35°20 ; , 144°40'. 39. 56°30', 236°30'. 41. 264°50', 275°10'. 43. 83°10', 263°10'. 45. 55°20', 235°20'. 47. 42°7', 222°7'. 49. 18°9', 341°51'. 51. 31°35', 328°25 ; . 53. 67°4', 247°4'. 55. 7°30', 187°30'. 57. 97°36', 262°24'. 59. 93°11', 273°11'. 61. 0.8016. 63. 3.079. 65. -100.00. 67. 0.3459. 69. -0.7073. 71. 1.214. 73. 0.220, 6.060. 75. 1.120, 4.260. 77. 0.755, 3.895. 79. 1.158, 5.122. 81. 1.143, 1.997. 83. 0.574, 5.706. EXERCISE 4-1. PAGE 96 27' - -v. ;. 81. 9. — < 64 1 1. 3y. 3. 5=. 5.100. 7.81. 9.—. 11. ay*. 13.1. Appendix B 343 15. x+2(xyy<* +y. 17. ^j-p 19. *"* ~ | - 21. 4(5*' 2 ). 28. (21)»'». 25# M!!!. 2?# (M^. 29.^. 31.^. 33.(5292)"*. 35. z™ y™. V x y 9 v ! u Q7 9 +3 \/2 QQ - (1 + yg) • x + V^^9 37. ? 39. 41. 43 ( x + y/x 2 - y 2 \ 2 i5 (3 - 2x) \/2x - x 2 < 4? 1 + z 2 - ^/l + x* 2 -x * s 2 (l + x 2 ) 49. 2(2 - z 2 ) 3 ' 2 . EXERCISE 4-2. PAGE 100 1. 54. 3. 3/2. 5. 6,652,800. 7. 5/24. 9. n(n - 1). 11. (n + l)n. 13. ^-^ • 15. — ^r • 19. x 7 - 7x« + 21a; 5 - 35x 4 + 35a; 3 - 21x 2 + 7x - 1. 21. 8a 6 - 36a 4 6 2 + 54a 2 6 4 - 276 6 . 23. x* 12 + 10a; 3 ' 2 + 40a; 1 ' 2 + SOar 1 ' 2 + 80ar 3 ' 2 + 32ar 6 ' 2 . 25.^-4^+6-4^+^. 27. x 12 - 6a; 10 ?/ 2 + 15z 8 ^ 4 - 20x«y« + 15a; 4 ?/ 8 - 6x 2 y 10 + ?/ 12 . 29. x 2 + ?/ 2 + z 2 + 2a;?y + 22/z + 2zz. 31. x 8 + 4a; 7 + 10a; 6 + 16a; 5 + 19a; 4 + 16a; 3 + 10a: 2 + 4x + 1. 5*. -8x 7 . 35. 7920a 8 6 4 . 37. -14a: 2 . 39. 2 11 • 3 5 • 5 • 7 • 13a; 33 ?/ 8 . 41. 924a: 3 ?/ 3 . EXERCISE 5-1. PAGE 104 1. log 2 8=3. 3. log 3 81 = 4. 5. log, 1000 - 3. 7. log 256 2 = i • 9. log 100 10 = 0.5. 11. y = log 10 x. 13. 8 2 = 64. 15. 2-« = ~ • 17. 7 3 = 343. 19. 10 4 = 10,000. 21. 4 3 ' 2 = 8. 23. 6. 25. 2. 4irr3 93 7 27. 3. 29. 1/10. 31. 5/2. 33. No solution. 35. log 6 ^- • 37. log5~r- 39. log* (u - y/u 2 - a 2 ). 41. (a) log* 7T + 1/2 log 6 Z - 1/2 log* g. (b) 2 log* 7 T + log* g - 2 log* W. EXERCISE 5-2. PAGE 107 1. 1. 3. 4. 5. -1. 7. -3. 9. -1. 11. 5. 13. -1. 15. -1. 17. 7314. 19. 7.314. 21. 7314000. 23. 0.007314. EXERCISE 5-3. PAGE 110 1. 1.5441. 3. 2.0212. 5. 7.7931-10. 7. 4.3636. 9. 9.5490-10. 11. 8.8215-10. 13. 9.9279-10. 15. 0.4536. 17. 7.8452-10. 19. 9.8908-10. 21. 0.4972. 23. 8.9439-10. 25. 9.9824-10. 27. 9.9476-10. 29. 9.1306-10. 31. 46.4. 33. 0.262. 35. 504. 37. 0.0000000000276. 39. 69.2. 41. 292.3. 43. 5,454,000,000. 45. 0.06114. 47. 4.554. 49. 0.00001072. 51. 0.6021. 53. 0.4266. 55. 1.585. 57. 3.728. 59. 2.5023. 61. 1.6297. 344 Appendix B EXERCISE 5-4. PAGE 112 1. 8.540. 3. 0.04292. 5. 3.183. 7. 0.0008416. 9. 0.1104. 11. 54.61. 13. 48.91. 15. 11,670. 17. 0.1795. 19. 0.02950. 21. 20.56. 23. 538,100. 25.1.708. 27.-1.021. 29.1.249. 31.0.4343,0.2171,9.5657-10,23.1,22.46. 33. 127,900,000 sq ft. 35. 12.62 ft. 37. 6,070,000 sq ft. 39. 16.5 amp. 41. $1,074.00. EXERCISE 5-5. PAGE 114 1. 2.3026. 3. 1.4429. 5. 0.8735. 7. 6.0001. 9. 1.5373. 11. 2.0794. 13. 1.4307. 15. 0.8228. EXERCISE 6-1. PAGE 119 1. B = 57°, b = 18, c = 22. 3. A = 18°50', a = 21, c = 66. 5. A = 27°!', B = 62°59', c = 7.012. 7. A = 22°44', b = 10.30, c = 11.17. 9. B = 53°39', b = 13.40, c = 16.64. 11. £ = 46°43', a = 73.66, c = 107.5. 13. A = 8°58', 5 = 81°2', c = 793.0. 15. A = 49°3', a = 2.663, c = 3.528. 17. h = 29 ft, I = 33 ft. 19. 113 ft. 21. 227 ft. 23. 13°34'. 25. 4 ft. EXERCISE 6-2. PAGE 124 1. 63°26', 11,000 ft. 3. 60°. 5. N11°24'W, 74 nautical miles. 7. 400 ft. 9. 80 ft, 173 ft. EXERCISE 6-3. PAGE 131 1. (a) 5. (c) 2 VS. (e) yft. (g) 4. 2. (a) 5 [3/5, 4/5]. <c) 2 V5 [- ^, ^] • (e)V^,^]. (g) 4 [0, 1]. 3. (a) 5, 53°8'. (c) 5, 143°8'. 4. (a) VS, 63°26'. (c) V73, 159°27'. 5. 12, 0°. 7. 22 knots, 20 knots. 9. 60 lb. 11. 49 lb, 250°45 / . EXERCISE 6-4. PAGE 133 I. 9.8733-10. 3. 8.8059-10. 5. 0.7391. 7. 9.3661-10. 9. 9.9427-10. II. 9.5906-10. 13. 0.0030. 15. 0.1004. 17. 1°20', 181°20'. 19. 8°30', 171°30'. 21. 18°, 198°. 23. 54°16', 234°16'. 25. 64°2', 295°58'. 27. 30°11', 210°11'. 29. 53°40', 233°40'. 31. 27°24', 152°36'. 33. 1°27', 181°27'. 35. 65°47', 294°13'. EXERCISE 6-5. PAGE 134 1. B = 59°, 6 = 60, c = 117. 3. B = 64°40', b = 133.9, c = 148.2. 5. A = 71°15', b = 2.01, a = 5.94. 7. A = 62°51', b = 18.53, c = 40.61. 9. 0°20'. 11. 671 lb, 200°40'. 13. 607 mph, N36°42'W. 15. 1815 lb, 1962 lb. 17. 49.19 in. 19. 16,900 lb. Appendix B 345 EXERCISE 7-1. PAGE 138 l.\(V6-V2). 3.f- 5. 2 -VS. 7. 2 -VS. •.-%$-£' 11. -3. 13. J (V5 -2 V2), J (1 + 2 a/6). 15. 4/5, -3/5. D O 17. 33/56, 63/16. EXERCISE 7-2. PAGE 142 I. i j/Tv5. 3. ^. 5. | ^2 + v^. 7. 1 + V& n / x 120 /ux V26 , . 120 , AS 119 , x 5\/26 , t . 119 »•(•)- 189. W^e", W-Tl9> W"W (e) ^6~' (f) l69' r \ ?o^ m^ 2856Q W 2197' W 239 II. (a) -3/5, (b) -^ |/50+5Vl0, (c) 3/4, (d) 4/3, (e)^|/50-5 v / 10, (f)-4/5, (g) ^~, 00 y- 13. cos 30. 15. tan 60. 17. sin 2 20. 19. cos 20. 21. tan 20. 23. cos 0. 25. cot £ • 27. tan* ~ • 4 2 EXERCISE 7-3. PAGE 144 1. i [sin 70 - sin 0]. 3. sin 100 + sin 20. 5. | [cos 60 + cos 20]. 7. _ I [cos 48° - cos 8°]. 9. - £ [cos 60 - cos 40]. 11. 2 sin ^ cos |. (\Q Of} 13. 2 sin y cos y • 15. - 2 sin 50° sin 30°. 17. 2 sin 32°30' cos 7°30'. 19. 2 sin 43° cos 3°. EXERCISE 8-1. PAGE 154 1. 2tt, 3, 0°. 3. 2tt, 1/2, 0°. 5. 8tt/3, 1/3, 0°. 7. 5tt/2, «, 0°. 9. 8 radians, 1, 0°. 11. 2?r/5, 3, 0°. 13. tt/6, «, 0°. 15. ir/4, «, 0°. 4 7 17. 1 radian, », — radians. 19. 2 radians, <» ; — radians. 21. 2x, 1, 0°. 23. 7r, 3, ^9 radian. EXERCISE 8-2. PAGE 161 I. s=?fiL±*. 3. s=^-=~. 5. x = -12t/ -22. 7. tt/6. 9. *r/6. II. ir/2. 13. tt/3. 15. tt/6. 17. -tt/3. 19. y . 21. -24°27'. 23. 12/5. 25. 5/13. 27. 3/4. 29. 0.3919. 31. ^ • 33. 3/4. 35. -3/5. 37. -tt/2. 39. -w. 41. w. 43. u. 45. Vl±Jf!. 47. Ut 49. w . 51. i/ u . 53. ^\ U \ - 55. \/T^uK 1 - ti 2 346 Appendix 6 EXERCISE 9-1. PAGE 168 1. x = 10/7, y = -6/7. 3. rr = -9/7, 2/ = 15/7. 8. a? = -2, */ = 1. f 18 - 1 o ~ 47 1 11 - 10 <w* 7 -* = 18 ,y= uT' 8 -* = "20"^ = 20 - ".a; =—,*/= 9/7. ** -I 1 68 ir T • j. i. iit - 21 T/rr ~22 13, a; = ~7F"; 2/ =: tH • 15. Inconsistent. 17. # = — — , y = 7/5, 2; = — £— • 19 - * =H * = -^ * =§' 21 * * = 10/7, y = 25/7, * = ~ 1/7 * 23. x = 16/11, y = -21/11, 2 = -9/2. 25. a = 9/5, y = -1/5, z = 1. 27. x = 10/7, y = 4, 2; = 20/3. 29. 3 = 0, 2/ = 5/7. 31. Inconsistent. 33. Inconsistent. 35. Consistent and dependent. 37. Consistent and independent. 39. Inconsistent. 41. Inconsistent unless c = 1. Consistent and dependent if c = 1. EXERCISE 9-2. PAGE 172 1. (1/4, 0), (0, -1/3). 3. (4, 0), (0, -4). 5. (0, 0). 7. (-4/3, 0), (0, 4). 9. (5/3, 0), (0, -5). 11. x = 6/5, y = 4/5. 13. x = 23/7, ?/ = 22/7. 15. x = 1, y = 5/8. 17. * = -j|,y=?- EXERCISE 10-1. PAGE 180 1. 5. 3. 0. 5. 0. 7. 7. 9. -11. 11. 22. 13. x =4.1, 1/ =0.3. 15. x = 25/13, ?/ = -5/13. 17. (y, 0), (0, -17). 19. (-3, -1), (2,4), (6, -3). EXERCISE 10-2. PAGE 185 1. x = 2, y = 3, z = -2. 3. x = 1, y = -2, z = 2/3. 5. a: = 2, */ = 3, z = —2. 7. x = — 8z, 1/ = -3z. 9. No nontrivial solution. EXERCISE 10-3. PAGE 187 1. 0. 3. 0. 5. -110. 7. 308. 9. 2184. EXERCISE 11-1. PAGE 193 I. + 4t, - 4i. 3. - 3r, + Si. 5. - 6\a\i, + 6|«|i. 7. 3 \/2 + 3 V§», 3 \/2 -3 V^'. 9. 1 + 4 \/2i, I - 4 V§*. II. y/\b ^ S\a\y/^ i, V^ -^\a\y/ob i. 13. -t. 15. -i. 17. -1. 19. f. 21. 1. 23. 0. 25. 0. 27. * = 3/2, y = 1/3. 29. x = 6, y = -5. 31. 3 = 1, y = -4. 33. x = 2, y = 5/2. 35. x = 7/3, y = 4/3. 37. 7 + 3t. 39. I + ^ i. 41. -1 + K. 43. 25 + Oi. 45. -1 + i. 47. 1 + Ot. Appendix B 347 49. -3 + (\/2 + y/2)i. 51. 6 + Oi. 53. 13 + lit. 55. -8 + 4i. 57. 11 - 3*. 59. 27 + 24*. 61. 28 + 16i. 63. 5 - 2 V§ + (2 + 5 y/2)i. 65. -2 - 26*. 67. ~ + I *. 69. ~ - -£■ *. 71. 33 - 22i. 73. ~ + £ i. J Z 41 41 Dl Dl 7 - 3 4. 77 025 1019 . 75- " 25 + 25 *' 77- 3233 + 3233 *" EXERCISE 11-2. PAGE 197 13. 3 - 2*. 15. -1 - 9*. 17. »2 + 3*. 19. 6 - 4*. 21. + 3*. 23. 5 + (2 - y/Z)i. 25. V2(cos 15° + i sin 45°). 27. 3(cos 270° + * sin 270°). 29. ~-(cos 54°44' + * sin 54°44'). 31. 13(cos 67 l 23' + i sin 67°23'). 33. 3 (cos 0° + * sin 0°). 35. ^p (cos 320°12' + * sin 320°12'). EXERCISE 11-3. PAGE 202 1. (i + VS) + (1 - VS)«. 8.-2V3+K. 5. (1 "" 2 V§) + (1+ 2 ^ 5) <. 7.1+1. 9.-1. 11. -1/2-^*. 13.^+^*. 16. 1 +|t. 17.-2"*. 19. — (0.6065 + 0.7951 *"). 21.-1. 23. 2[cos (9° + A' 90°) + * sin (9° + A; 90°)], ft = 0, 1, 2, 3. 25. 2[cos (36° + ft 72°) + i (30° + ft 72°)], ft = 0, 1, 2, 3, 4. 27. 1, i, -1, -t. 29. cos(S0° + ft 120°) + i sin (80° + ft 120°), ft = 0, 1, 2. Q1 14 + 18* QQ 828 + 154* , 31. amperes. 33. ~=^ ohms. o 17o EXERCISE 12-1. PAGE 206 1. 0, -7. 3. -1/2, -3/4. 5. 3, -2. 7. 3,3, -3. 9. sin = 0, 1; 0°, 90°, 180°. 11. sin = 1, -2; 90°. 13. sec 0=2, -8; 00°, 300°, 97°11', 262°49'. 15. cot = -3, 2; 161°34', 341°34', 2(>°34', 206°34'. 17. cot 0=7, -17, 8°8', 188°S', 17G°38', 356°38'. 19. sin = 0, 2, - | ; 0°, 180°. EXERCISE 12-2. PAGE 209 I. 10, -2. 3. 10, -3. 5. - 1= ^ . 7. ^if^i. 9. tan = 1 ± V5; « = 67°30 ; , 247°30', 157°30', 337°30 / . II. sec = 1, -3; 0°, 109°28', 250°32 / . 348 Appendix B 13. esc 6 = * ± 4 V ^ ; 36 Q 23', 143°37', 237°30', 302°30'. 15. cos $ = 0.4142, -2.414, 65°32', 294°28'. 17. (a - 3) 2 + 4(y + 2) 2 = 4. 19. (s - 5) 2 + My - 5)2 = 16. 21. 4(s - 2) 2 + 9(i/ - l) 2 = 36. 23. 4(rc + 4) 2 - 9(y - 2) 2 = -36. 25. 3(x + 10/3) 2 - (y + l) 2 = 64/3. 27. \/2[(s -4) 2 +9/2]. 29. . 31. [2((rc + 7) 2 - 32)]~" 2 . y/{z — 3) 2 - 16 33. [9((a; + 4/3) a + l)]-i/3. EXERCISE 12-3. PAGE 212 I. 1, -7. 3. -1 ± >/6. 5. -1, -1. 7. - 3 ^ v7i < 9 5/2 ^ _ 1/2> II. -1, - 11/8. 13. -1, -15/7. 15. tan = 1, -5/2, 45°, 225°, 111°48', 291°48'. 17. sin 6 = 3.7913, -0.7913; 232°18', 307°42'. 19. cos = r ; no values of 0. o 21. sec = 0.414, -2.414; 114°28', 245°32'. 23. esc 6 = -0.6972, -4.3028; 193°26', 346°34'. EXERCISE 12-4. PAGE 213 1. 18. 3. -4. 5. 8 ± 4 \/5. 7. 5. 9. *izi|-^Jl . n # l, 5. EXERCISE 12-5. PAGE 215 1. d= a/5, ±2i. 3. ±», ±t VO. 5. ±2, ±3. 7. 1 ^ 2l • 9. -4 ± VT5, 3±2y^. EXERCISE 12-6. PAGE 216 1. Conjugate imaginary. 3. Heal, unequal, irrational. 5. Real, unequal, irrational. 7. Real, unequal, rational. 9. Real, equal, rational. 11. Conjugate imaginary. 13. Real, unequal, irrational. 15. Conjugate imaginary. EXERCISE 12-7. PAGE 218 1.-2,-1. 3.0,2. 5.3/2,6/5. 7.6/5,-1/5. 9. |> ~~p 11. a* - 2x = 0. 13. a 2 - 9x + 18 = 17.3 2 + 1=0. 19. z 2 - (\^ - Vl)x = 0. 21. z 2 ~2V§z+8 = 0. '•f'lffiT 11. » 2 -2x=0. 13. a; 2 - 9x + 18 = 0. 15. s 2 - 4a; + 9 = 0. Appendix B 349 EXERCISE 12-8. PAGE 224 1. Circle. 3. Intersecting lines. 5. Hyperbola. 7. Parabola. 9. Intersecting lines. 11. Hyperbola. 13. Intersecting lines. 15. Intersecting lines. 17. Intersecting lines. 19. Parallel lines. EXERCISE 12-9. PAGE 227 1. (3, 3), (-3/2, 3/4). 3. (3, 4), (-4, -3). 5. (2, 4), (-3, 9). 7. (3, 2), (-1, -6). 9. (2, 3). 11. (± 2 V§, 1). 13. No solution. 15. (±4, ±2). EXERCISE 12-10. PAGE 231 1. (3, 3), (-3/2, 3/4). 3. (3, 4), (-4, -3). 5. (4, -3), (-2, 6). 7. (4, 0), (-5, 3). 9. (4, 6), (-3, -1). 11. (±3, I) (±3 yAdi, -10). 13. (*^;,*^). 15. (±2, ±4). 17. (1,2), (2,1). 19. (4,1), (-4,-1), (14, -4), (-14,4). 21. (4, 2), (-4, -2), (V6, -2 V6,), (- V6, 2 ^6). 23. (5, 5), (-5, -5). 25. (3, 4) (4, 3), (-3, -4), (-4, -3). 27. (2 + i V2, 2 - i y/2), ( -3 + i \fl, -3 - i y/l), (2 - i \/2 y 2 + % y/2), (-3 -iV7, -3 +iV7). / 31 -3 V93 31+3 \/93 \ / 31 + 3 \/93 31-3 \/93 \ ^- v 31 ' 31 ; ' v 31 ' 3i~; * EXERCISE 12-11. PAGE 233 1. 6. 3. 4/3. 5. 2.292. 7. 1.682. 9. 1. 11. 1.836. 13. 2.718. 15. 8.547. 17. 2.944. 19. 49.3. 21. 10, 0.1. 23. 0.7S74. 25. -0.44. 27 logO -ac) -log 6- log c , 2d ^ {y±v - r — ih , La8M4> log c EXERCISE 13-1. PAGE 239 I. Q : x - 7, # : 0. 3. Q : x - 1, 72 : 2. 5. Q : x 3 - 3.r 2 + 6x - 24, # : 78. 7. Q : 2z 3 + 3x 2 + 4, 72 : 0. 9. Q : x 2 + 5x + 8, R : 11. II. Q : x 2 + 2x - 15, R : 0. 13. Q : a; 3 + 3a; - 6, R : 0. 15. Q : x**- 1 + x*~ 2 y + • • • + y n ~ l , R : 0. 17. 52, 2. 19. 24, -36. EXERCISE 13-2. PAGE 241 1. 2a; 2 - 2x - 3. 3. Not a factor. 5. x 2 -f 2x - 5. 7. x 7 + 2xHj 2 + 4.rV -J- 8*V + 16x*y* + 32x 2 y 10 + 64.ri/ ,a -f 128y". 9. Not a factor. 11. hx 2 + 2ab(5 - Sab 2 )x - 12a 3 6 4 . 13. Not a factor. 15. 12x 3 - 22a: 2 - 34a; + 60. 17. 24a; 3 - 90a; 2 + 39a; + 45. 350 Appendix B EXERCISE 13-3. PAGE 244 1. 1 ± V&, -3. 3. 1 =fc i, ±3. 5. 2 ± V2i, 1, 1, 2. 7. s 3 - ix 2 + 9x - 10 = 0. EXERCISE 13-4. PAGE 247 1. -1/3, 1/2, 5/3. 9. -1. 11. 2, 2, -2, -2. 13. 2/3, ~* y^ - 15. 1/2, =bi. 17. 3/5, 1 =fc i. 19. 1/2, ± V5- EXERCISE 14-1. PAGE 253 I. x < 3. 3. a: > -5. 5. x < 4. 7. a; < -1/2. 9. x < -1. II. -\<*<\- M.-§<*<|« 15 --§ <a;< jr 17. ~13<x<13. 19. -4 £ x £ 4. 21. -1 < x < 1/3. 23. x < -5/3, x > 2. 25. No values of s. 27. x < -3, -2 < x < -1. 29. 1 < a < 2, a; > 3. 31. | x | ^ 5. 33. x < -2, s > 0. 35. 3 < -1/2. EXERCISE 15-1. PAGE 260 1.1,2,4,7,11. 3.1,2,3,5,11,35. 7. 2, 5, 11, 23, 47; 2, 7, 18, 41, 88. EXERCISE 15-2. PAGE 263 1. 17, 20. 3. 18, 25. 5. Not an arithmetic progression. 7. 46 — 3a, 5b — 4a. 9. 4fl = 26 , 5a ~ 36 . 11. f 26 = 78, £ 26 = 1053. 13. Iio = 10, S i0 = 55. 15. hi = 149, <S 75 = 5625. 17. J 20 = 5.9, S 20 = 61. 19. i,oo = 100, S, oo =505O. 21. l l0 = 100, S,o = 550. 23. a = - y , i 4 « = y • 25. a = 1, n = 13 or a = -1/2, n = 16. 27. /« = 9, n = 9. 29. n 2 . 31. d = -^°- • 33. 33, 16. 35. d = - ~* • 37. $37.75. 39. 282. 11 ' a(k + 1) EXERCISE 15-3. PAGE 264 1. 1/15, 1/19. 3. 1/20, 1/25. 5. 4, 16/5. 7. 1/47. 9. j, | , y , |. „ 120 60 120 , TT'T'TT' EXERCISE 15-4. PAGE 267 I. 128, 512. 3. 256, 1024. 5. 8, 16/3. 7. ^, ^. 9. -1, 1. II. l„ = P , 5„ = 9[1 - (2/3)"]. 13. J 101 = 10-", 5 101 = yp [1 + 10—>]. 15./.=^,^=^^. 17.3,-14. 19.^51. 21. ±6, 15/2. Appendix B 351 23. 10; 100; 1000; 10,000; 100,000. 1 *■-• x n fix n 27. As n increases, the sum approaches 3 as a limit. 29. (1 - a;) 2 1 - x EXERCISE 15-5. PAGE 270 1.64/65. 3. 9 + 2 5# 16 ' 7# 3 - 9 - 5/8 - 1L 2/U - 13. 36, 36 [1 - (1/3) 20 ], 36(1/3)20. 15. 12 ft, 12 [1 - (3/4) 10 ] feet. 17. 1/9. 19.10/11. 21.1/6. 23.3. 25. g^- EXERCISE 15-6. PAGE VT2 1.1-*+-—. 3 -i+l"f+S- 8 - 1 -5 + X-S- 7 -i-S+S"S- «-i-S + SP-^- »• 1-2190. 13.1.3684. X X 2 X 3 X* X 3 X* X* x° 15. 0.8508. 17. 0.9415. 19. 1.0149. EXERCISE 17-1. PAGE 279 1. 72. 3. 216. 5. 9999. 7. 20,160; 7560. EXERCISE 17-2. PAGE 282 1. 20; 210; 95,040; 143,640; 970,200. 3. (a) 720, (b) 48, (c) 480. 5. 125. 7. 8 Co<z 8 4- sCitfb + 8 C 2 a 6 6 2 + 8 C 3 a*b* + 8 C 4 a<6 4 + 8 C 5 a 3 6 6 + 8 C 6 a 2 6« + sC 7 ab 7 -f 8 C 8 6 8 . 9. 9,979,200. 11. 66. 13. 63. 15. 52 " (13!) 4 EXERCISE 17-3. PAGE 288 1. 2/5, $24.00. 3. 1/3, 2/3. 5. $7.29. 7. $2.42. 9. 0.553. 11. 16/63, iQ JL 1* A 215 16. 1728 - 15. 72 , 216 ' EXERCISE 18-1. PAGE 295 1. C = 9S°5S', 6 = 14.55, c = 20.46. 3. A = SOW, a = 1169, 6 = 1079. 5. B = 5°31', 6 = 1.051, c = 7.513. 7. A = 17°3', £ = 100°26', b = 71.18. 9. B = 24°27', C = 101 °20', c = 1193. 11. No solution. 13. 615.3 ft. 15. 449 ft. 17. 12.6 in. 19. 3158 lb, 101°37'. EXERCISE 18-2. PAGE 298 I. c = 270, B = 51°30', A = 69°50'. 3. a = 290, B = 11°50', C = 101°. 5. a = 100, £ = 10°, C = 20°. 7. 48°20'. 9. No solution. II. A = 54°20', £ = SOW, C = 66°. 13. 3257 ft. 15. 700 ft. 17. c = 20, A = 63°, B = 73°, C = 44°. 352 Appendix 8 EXERCISE 18-3. PAGE 301 1. A = 50°, B = 70°, c = 56. 3. A = 61°, C = 23°50', 6 = 262. 5. B = 31°11', C = 70°, a = 79.25. 7. A = 47°, £ = 58°, C = 75°. 9. A = 42°20', B = 57°30', C = 80°10'. 11. 4 in., 5 in. 13. 11 in. 15. 55°30', 59°50', 64°40'. EXERCISE 18-4. PAGE 304 1. 88.41. 3. 243,900. 5. 9285. 7. 90. 11. 54 ft. 15. 45 ft. 17. 38 sq ft, 12 sq ft. Index Abscissa, 50 Absolute value, 14, 57 Absolute value function, 57 Addition, 2 of algebraic expressions, 21 of fractions, 36 fundamental laws for, 2 of real numbers, 2 Addition formulas, 135 Algebraic expressions, 17 addition and subtraction of, 21 division of, 23 multiplication of, 22 symbols of grouping of, 18 transposing terms of, 43 Amplitude of a trigonometric function, 149, 152 Angle coterminal, 76 definition of, 75 degree measure of, 77 of depression, 120 of elevation, 120 initial side of, 75 measurement of, 76 negative of, 75 positive of, 75 radian measure of, 77 standard position of, 75 terminal side of, 75 trigonometric functions of, 81, 82 vertex of, 75 Antilogarithms, 109 Arc of a circle, 78 Area of a sector of a circle, 78 of a triangle, 302 Arithmetic means, 262 Associative law for addition, 2 for multiplication, 2 Axes, coordinate, 49 Axis of symmetry of a parabola, 219 Base change of logarithmic, 113 exponent and, 16, 86 of logarithms, 101, 113 Bearing in navigation and surveying, 121 Binomial, 17 Binomial series, 271 Binomial theorem, 97 combinations applied to, 282 general term of, 99 proof of, for positive integral exponents, 275 Braces, 18 Brackets, 18 Cartesian coordinates, 50 Characteristic of a logarithm, 106 Characteristic of the general quadratic equation, 222 Circular system, 77 Classification of functions, 61 Coefficient, 17, 18 Combinations, 278, 281 and the binomial coefficients, 282 Common difference, 260 Common logarithms, 105 Common ratio, 265 Commutative law, 2 Completing the square, 206 Complex fractions, 40 Complex numbers, 189 absolute value of, 196 addition and subtraction of, 192 amplitude of, 196 argument of, 196 congugate of, 191 definition of, 189 De Moivre's Theorem, 199 division of, 193, 198 graphical representation of, 195 modulus of, 196 multiplication of, 192, 198 roots of, 200 trigonometric representation of, 196 Composite number, 25 Composite (reducible) polynomial, 26 Conditional equation, 18 Constant, 53 Constant function, 53 Convergence of series, 259 Coordinate systems Cartesian, 50 one-dimensional, 49 rectangular, 49 Coordinates, 49, 50 Cosecant, definition of, 65 353 354 Index Cosine, definition of, 64 Cotangent, definition of, 65 Coterminal angles, 76 Counting numbers, 1 Defective equations, 45 Degree of a polynomial, 17 relation to radian, 77 of a term, 17 term of highest, 17 unit of angle measure, 77 De Moivre's Theorem, 199 Dependent events, 286 Descartes, Rene, 50 Determinants, 173 expansion of, 176 minors and cof actors of, 175 principal and secondary diagonal of, 174 properties of, 177 of the second order, 173 solution of linear equations by means of, 181, 183, 184 sum and product of, 185 of the third order, 175 Directed line segment, 122 Discriminant, 215 Distance between points, 50, 51 Distributive law, 3 Division, 5, 16, 23, 25, 39 Domain of a function, 53 Double-angle formulas, 139 Empirical probability, 284 Equality symbol, 12 Equations conditional, 18 consistent, 163, 164, 165 defective, 45 definition of, 18 dependent, 165 equivalent, 43 extraneous roots of, 45, 213 inconsistent, 163, 165 independent, 164 linear, 42, 43, 163, 166, 167 in quadratic form, 204, 214 solution of, 42, 163, 171, 181, 183, 184, 212, 224, 227 Exponential equations, 231 Exponents, 16, 86, 88, 90, 92 Factor theorem, 240 Factorial symbol, 97 Factoring, 25, 27 Fractions, 9 addition and substraction of, 36 Fractions (Conh) complex, 40 fundamental operations on, 9, 10 multiplication and division of, 39 reduction of, 33 signs associated with, 34 Functions absolute value, 57 algebraic, 61 classification of, 61 constant, 53 definition of, 53 domain of, 53 exponential, 233 graphs of, 146, 147, 148, 169, 218, 233, 244 greatest integer, 57 identity, 53 inverse, 155 irrational, 62 linear, 54, 169 logarithmic, 233 multiple-valued, 53 notation for, 55 periodic, 149 point, 63 polynomial, 61 range of, 53 rational, 61 rule of correspondence of, 53 single-valued, 53 transcendental, 61 trigonometric, 63 Fundamental assumptions, 1 Fundamental operations, 1, 9, 10, 20 Fundamental theorem of algebra, 241 General term in the binomial expansion, 99 Graphs of exponential functions, 233 of inverse trigonometric functions, 158, 159 of linear functions, 169 of logarithmic functions, 233 of polynomials for large values of x, 244 of trigonometric functions, 146, 147, 148, 149, 151, 152 Greater-than symbol, 12 Greatest common divisor, 30 Greatest-integer function, 57 Half -angle formulas, 140, 300 Highest common factor, 31 Highest degree term, 17 Horizontal line, 50, 169 Identity, 18 Independent events, 286 Index 355 Inequalities, 248 absolute, 248, 254 conditional, 248, 249 involving absolute values, 14 properties of, 248 solution of conditional, 249 Inequality symbol, 12 Initial side, 75 Intercepts, 170 Interpolation, 83, 108 Inverse functions, 155 Irrational functions, 62 Irrational number, 1 Irreducible polynomial, 26 Law of cosines, 296 Law of sines, 289 Law of tangents, 298 Law(s) of exponents, 86 Least common multiple, 32 Less-than symbol, 12 . Like terms, 21 Limit of sequence, 258, 259 Line horizontal, 50, 169 vertical, 51, 169 Linear equation, 42 graphs of, 163, 169, 171 in one unknown, 43 Literal parts, 17 Logarithmic computation, 110 Logarithmic equations, 231 Logarithms, 101 base of, 101 change of base, 113 characteristic of, 106 common, 105 computation by, 110 definition of, 101 laws of, 102 mantissa of, 106, 108 Napierian (natural), 105 tables of, 108 of trigonometric functions, 131 Mantissa of a logarithm, 106 Mathematical expectation, 284 Mathematical induction, 273 Mean arithmetic, 262 geometric, 267 harmonic, 264 Meaning of a m/n , 94 Meaning of a , 98 Measurement of angles, 76 Monic polynomial, 18 Most probable number, 284 Multinomial, 17 Multiplication of algebraic expressions, 22 of fractions, 39 fundamental laws for, 2 Mutually exclusive events, 285 Natural (Napierian) logarithms, 105 Negative exponents, 90 Negative numbers, 7, 8 Number algebraic, 62 complex, 189 irrational, 1 negative, 1, 7, 8 positive, 1, 7, 8 prime, 25 real, 1 transcendental, 62 Number scale, 6, 49 One-to-one correspondence, 49 Operations on fractions, 9, 10 with zero, 5 Order of fundamental operations, 20 Order relations for real numbers, 12 Ordered number pairs, 50 Ordinate, 50 Origin, 6, 49 Parentheses, 18 Period, 150 Periodicity, 149, 150 Permutations, 278, 279 Phase, 149, 152 Point function P (t), 63 Polynomial, 17, 18, 26, 61 Positive integer, 1 Positive integral exponents, 16, 8G Positive numbers, 7, 8 Power, 16, 17, 86, 87, 88 Prime, 25, 26 Principal branch, 158 Principal value, 158, 159, 160 Probability, 278 empirical, 284 mathematical, 283 Product, definition of, 2 Product formulas, 143 Progressions, 256 arithmetic, 260 geometric, 265 armonic, 264 infinite geometric, 268 356 Index Projections, 122 Quadrant, 49 Quadratic equations methods for solving, 204, 209 in one unknown, 204 in two unknowns, 221, 224, 227 Quotient, 5 Radian, 77 Radius vector, 52 Range of a function, 53 Rational exponents, 92 Rational function, 61 Rational number, 1 Real number, 1 Reciprocal, 4, 5 Rectangular coordinates, 49, 50 Redundant equation, 45 Remainder theorem, 240 Repeated trials, 287 Repeating decimals, 269 Root of an equation, 42, 45, 213, 217 Scalar quantities, 125 Scientific notation, 92 Secant, definition of, 65 Second degree equation, 204 Segment, line, 122 Sequences, 256, 258, 259 Series, 256, 257, 259 Sexagesimal system, 77 Simultaneous equations, 163, 171, 181, 224, 227 Sine, definition of, 64 Special products, 22 Standard position of an angle, 75 Sum, definition of , 2 Symbols of grouping, 18 Synthetic division, 235 Tables of logarithms, 108 of trigonometric functions, 71, §2 Tangent, definition of, 64 Terminal side, 75 Terms of algebraic expressions, 2, 17 Theory of equations, 235 Transcendental functions, 62 Triangles, 115 solution of general, 289 solution of right, 117, 133 Trigonometric functions, 63 of angles, 81, 116 definitions of, 64, 65 graphs of, 146, 147, 148, 149, 151, 152 of important special numbers, 65 inverse, 146, 156, 158, 159, 160 logarithms of, 131 of sums and differences, 135 variation of, 146 Trigonometric identities, 68 Variable, 53 dependent, 53 independent, 53 Variation, 57, 58 Vector, 125 components of, 125 magnitude of, 126 multiplication of, by a scalar, 126 normalization of, 127 projection of, 125 representation of, 125, 128 sums and differences of, 127, 129 Vertex of an angle, 75 Vertical lines, 51, 169 a?-axis, 49 x-coordinate, 50 y-axis, 49 ^/-coordinate, 50 Zero, 3, 5 Zero polynomial, 61